## Trigonometry – ICSE Solutions for Class 10 Mathematics

Get ICSE Solutions for Class 10 Mathematics Chapter 18 Trigonometry for ICSE Board Examinations on APlusTopper.com. We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. You can download the Class 10 Maths ICSE Textbook Solutions with Free PDF download option.

### Determine the Following

Question 2. Without using tables evaluate

Question 4. Without using trigonometric tables, evaluate

Question 9. From trigonometric tables, write the values of:

Question 10. The string of a kite is 150 m long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.

Question 11. Solve the following equations:

Question 12. Using trigonometric tables evaluate the following:

### Figure Based Questions

Question 1. In figures, find the length CF.

Question 2. With reference to the figure given alongside, a man stands on the ground at a point A, which is on the same horizontal plane as B, the foot of a vertical pole BC. The height of the pole is 10 m. The man’s eye is 2 m above the ground. He observes the angle of elevation at C, the top of the pole as x°, where tan x° = 2/5.

Question 3. From the top of a tower 60 m high, the angles of depression of the top and bottom of pole are observed to be 45° and 60° respectively. Find the height of the pole.

Question 4. In triangle ABC, AB = 12 cm, LB = 58°, the perpendicular from A to BC meets it at D. The bisector of angle ABC meets AD at E. Calculate:
(i) The length of BD;
(ii) The length of ED.

Question 5. From the top of a light house 100 m high the angles of depression of two ships on opposite sides of it are 48° and 36° respectively. Find the distance between the two ships to the nearest metre.

### Concept Based Questions

Question 2. From a light house, the angles of depression of two ships on opposite sides of the light house were observed to be 30° and 45°. If the height of the light house is 90 metres and the line joining the two ships passes through the foot of the light house, find the distance between the two ships, correct to two decimal places.

Question 3. A man on the deck of a ship is 10 m above water level. He observes that the angle of elevation of the top of a cliff is 42° and the angle of depression of the base is 20°. Calculate the distance of the cliff from the ship and the height of the cliff.

Question 4. A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.

Question 5. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7 metres. At a point in a plane the angle of elevation of the bottom and the top of the flagstaff are respectively 30° and 60°. Find the height of the tower.

Question 6. A pole being broken by the wind the top struck the ground at an angle of 30° and at a distance of 8m from the foot of the pole. Find the whole height of the pole.

Question 7. The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.

Question 8. A man on the top of vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower ? (Give your answer correct to nearest seconds).

Question 9. Two men on either side of a temple 75 m high observed the angle of elevation of the top of the temple to be 30° and 60° respectively. Find the distance between the two men.

Question 10. An aeroplane when 3,000 meters high passes vertically above another aeroplane at an instance when their angles of elevation at the some observation point are 60° and 45° respectively. How many meters higher is the one than the other.

Question 11. From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places.

Question 12. A man is standing on the deck of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of hill.

Question 14. Vertical tower is 20m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?

Question 15. Two person standing on the same side of a tower in a straight line with it measure the angle of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70 m find the distance between the two person.

Question 16. As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the light house in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.

Question 17. An aeroplane at an altitude of 250 m observes the angle of depression of two Boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution: Let the width of the river CD be x,

Question 19. (i) The angle of elevation of a cloud from a point 200 metres above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud.
(ii) If the angle of elevation of a cloud from a point h meters above a lake is a*and the angle of depression of its reflection in the lake is |i. Prove that the height of the cloud is

Question 20. From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.

Question 21. A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m., away from the bank, he finds the angle of elevation to be 30°. Calculate:

For More Resources

## Trigonometric Ratios Of Some Specific Angles

The angles 0°, 30°, 45°, 60°, 90° are angles for which we have values of T.R.

 θ 0° 30° 45° 60° 90° Sin 0 1/2 1/√2 √3/2 1 Cos 1 √3/2 1/√2 1/2 0 Tan 0 1/√3 1 √3 ∞ Cosec ∞ 2 √2 2/√3 1 Sec 1 2/√3 √2 2 ∞ Cot ∞ √3 1 1/√3 0

## Trigonometric Ratios Of Some Specific Angles With Examples

Example 1:    Evaluate each of the following in the simplest form:
(i) sin 60º cos 30º + cos 60º sin 30º
(ii) sin 60º cos 45º + cos 60º sin 45º
Sol.      (i)  sin 60º cos 30º + cos 60º sin 30º
$$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=1$$
(ii)  sin 60º cos 45º + cos 60º sin 45º
$$=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}$$
$$=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$$

Example 2:     Evaluate the following expression:
(i) tan 60º cosec2 45º + sec2 60º tan 45º
(ii) 4cot2 45º – sec2 60º + sin2 60º + cos2 90º.
Sol.      (i)  tan 60º cosec2 45º + sec2 60º tan 45º
= tan 60º (cosec 45º)2 + (sec 60º)2 tan 45º
= √3 × (√2)2 + (2)2 × 1
= √3  × 2 + 4 = 4 + 2 √3
(ii)  4cot2 45º – sec2 60º + sin2 60º + cos2 90º
= 4(cot 45º)2 – (sec 60º)2 + (sin 60º)2 + (cos 90º)2
= 4 × (1)2 – (2)2 + (√3/2)+ 0
= 4 – 4 + 3/4 + 0 = 3/4

Example 3:     Show that:
(i) 2(cos245º + tan260º) – 6(sin245º – tan230º) = 6
(ii) 2(cos460º + sin430º) – (tan260º + cot245º) + 3 sec230º = 1/4
Sol.    (i)  2(cos245º + tan260º) – 6(sin245º – tan230º)
$$=2\left( {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{(\sqrt{3})}^{2}} \right)-6\left( {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \frac{1}{\sqrt{3}} \right)}^{2}} \right)$$
$$=2\left( \frac{1}{2}+3 \right)-6\left( \frac{1}{2}-\frac{1}{3} \right)=2\left( \frac{1+6}{2} \right)-6\left( \frac{3-2}{6} \right)$$
$$=2\times \frac{7}{2}-6\times \frac{1}{6}=7-1=6$$
(ii)  2(cos460º + sin430º) – (tan260º + cot245º) + 3 sec230º
$$=2\left( {{\left( \frac{1}{2} \right)}^{4}}+{{\left( \frac{1}{2} \right)}^{4}} \right)-\left( {{(\sqrt{3})}^{2}}+{{(1)}^{2}} \right)+3{{\left( \frac{2}{\sqrt{3}} \right)}^{2}}$$
$$=2\left( \frac{1}{16}+\frac{1}{16} \right)\left( 3+1 \right)+3\times \frac{4}{3}$$
$$=2\times ~\frac{1}{8}-4+4=\frac{1}{4}$$

Example 4:     Find the value of x in each of the following :
(i) tan 3x = sin 45º cos 45º + sin 30º
(ii) cos x = cos 60º cos 30º + sin 60º sin 30º
Sol.    (i)  tan 3x = sin 45º cos 45º + sin 30º
$$\Rightarrow tan\text{ }3x=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+\frac{1}{2}$$
$$\Rightarrow tan\text{ }3x=\frac{1}{2}+\frac{1}{2}$$
⇒ tan 3x = 1
⇒ tan 3x = tan 45º
⇒ 3x = 45º    ⇒      x = 15º
(ii)  cos x = cos 60º cos 30º + sin 60º sin 30º
$$\Rightarrow cos\text{ }x=\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}$$
$$\Rightarrow cos\text{ }x=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$$
⇒ cos x = √3/2
⇒ cos x = cos 30º
⇒ x = 30º

Example 5:     If x = 30°, verify that
$$(i)\tan 2x=\frac{2\tan x}{1-{{\tan }^{2}}x}\text{ }(ii)\sin x=\sqrt{\frac{1-\cos 2x}{2}}$$
Sol.(i)   When x = 30°, we have 2x = 60° .
∴  tan 2x = tan 60° = √3
$$\text{And, }\frac{2\tan x}{1-{{\tan }^{2}}x}=\frac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }$$
$$=\frac{2\times \frac{1}{\sqrt{3}}}{1-{{\left( \frac{1}{\sqrt{3}} \right)}^{2}}}$$
$$=\frac{2/\sqrt{3}}{1-\frac{1}{3}}=\frac{2/\sqrt{3}}{2/3}=\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}$$
$$\tan 2x=\frac{2\tan x}{1-{{\tan }^{2}}x}$$
(ii)        When x = 30°, we have 2x = 60°.
$$\sqrt{\frac{1-\cos 2x}{2}}=\sqrt{\frac{1-\cos 60{}^\circ }{2}}$$
$$\sqrt{\frac{1-\frac{1}{2}}{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$$
And, sinx = sin30° = 1/2
$$\sin x=\frac{\sqrt{1-\cos 2x}}{2}$$

Example 6:     Find the value of θ in each of the following :
(i) 2 sin 2θ = √3      (ii) 2 cos 3θ = 1
(i)  2 sin 2θ= √3
⇒ sin 2θ= √3/2
⇒ sin 2θ= sin 60°
⇒ 2θ= 60°   ⇒    θ = 30°
(ii)  2 cos 3θ = 1
⇒ cos 3θ = 1/2
⇒ cos 3θ = cos 60°
⇒ 3θ = 60°   ⇒   θ = 20°.

Example 7:    If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2θ + sin2θ – 1.
Sol. sin θ = cos θ
$$\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{\cos \theta }{\cos \theta }$$
[Dividing both sides by cos θ]
⇒ tanθ = 1
⇒ tanθ = tan45°  ⇒  θ= 45°
∴ 2 tan2θ + sin2θ – 1
= 2tan245° + sin245° – 1
$$=2{{(2)}^{2}}+{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-1$$
$$=2+\frac{1}{2}-1=\frac{5}{2}-1=\frac{3}{2}$$

Example 8:     An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side.
Sol.    Let ABC be an equilateral triangle inscribed in a circle of radius 6 cm. Let O be the centre of the circle.

Then, OA = OB = OC = 6 cm.
Let OD be perpendicular from O on side BC. Then, D is mid-point of BC and OB and OC are bisectors of ∠B and ∠C respectively.
∴ ∠OBD = 30°
In ∆OBD, right angled at D, we have
∠OBD = 30° and OB = 6 cm.
$$\cos \angle OBD=\frac{BD}{OB}\Rightarrow \cos {{60}^{0}}=\frac{BD}{6}$$
$$\Rightarrow BD=6\cos {{60}^{0}}=6\times \frac{\sqrt{3}}{2}=3\sqrt{3}\text{ }$$
⇒ BC = 2 BD = 2(3√3 )cm = 6 √3 cm.

Example 9:     Using the formula, sin(A – B) = sinA cosB – cosA sinB, find the value of sin 15º.
Sol.    Let A = 45º and B = 30º. Then A – B = 15º. Putting A = 45º and B = 30º in the given formula, we get
sin(45º – 30º) = sin45º cos30º – cos45º sin30º
$$or,\sin (45{}^\text{o}-30{}^\text{o})=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$$
$$=\frac{\sqrt{3}-1}{2\sqrt{2}}\Rightarrow \sin 15{}^\text{o}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$

Example 10:     If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90° ; A > B, find A and B.
tan (A + B) = √3 = tan 60° & tan (A – B) = 1/√3 = tan 30°
A + B = 60° …….(1)
A – B = 30° …….(2)
2A = 90°  ⇒  A = 45°
A + B = 60
A – B = 30
Subtract equation (2) from (1)
A + B = 60
A – B = 30
2B = 30°
⇒ B = 15°. Ans.
Note: sin(A + B) = sin A cos B + cos A sin B
sin(A + B) ≠ sin A + sin B.

## Heights And Distances

Angle Of Elevation
The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e. the case when we raise our head to look at the object. (see fig.)

Angle Of Depression
The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e. the case when we lower our head to look at the point being viewed. (See fig.)

## Heights And Distances With Examples

Example 1:    The shadow of a building is 20 m long when the angle of elevation of the sun is 60º. Find the height of the building.
Sol.    Let AB be the building and AC be its shadow.

Then, AC = 20 m and ∠ACB = 60º.
Let AB = h.
Then, $$\frac { AB }{ AC }$$ = tan 60º = √3
⇒ h/20 = √3
∴ h = (20 × √3 )m = (20 × 1.732) m
= 34.64 m.

Example 2:    If a vertical pole 6m high has a shadow of length 2 √3 metres, find the angle of elevation of the sun.
Sol.    Let AB be the vertical pole and AC be its shadow.

Let the angle of elevation be θ. Then,
AB = 6 m, AC = 2 √3 m
and ∠ACB = θ.
$$\text{Now, tan }\theta =\frac{AB}{AC}=\frac{6}{2\sqrt{3}}=\sqrt{3}=\text{ tan }60{}^\text{o}.$$
∴ θ = 60º.

Example 3:    A ladder against a vertical wall makes an angle of 45º with the ground. The foot of the ladder is 3m from the wall. Find the length of the ladder.
Sol.    Let AB be the wall and CB, the ladder.

Then, AC = 3m and ∠ACB = 45º
$$\text{Now, }\frac{CB}{AC}=\sec \text{ 45}{}^\text{o}=\sqrt{2}\Rightarrow \frac{CB}{3}=\sqrt{2}$$
∴  Length of the ladder = CB = 3 √2
= (3 × 1.41) m = 4.23 m

Example 4:    A balloon is connected to a meteorological station by a cable of length 200 m, inclined at 60º to the horizontal. Find the height of the balloon from the ground. Assume that there is no slack in the cable.
Sol.    Let B be the balloon and AB be the vertical height. Let C be the meteorological station and CB be the cable.

Then, BC = 200 m and ∠ACB = 60º
$$\text{Then, }\frac{AB}{BC}=\text{sin }60{}^\text{o}=\frac{\sqrt{3}}{2}$$
$$\Rightarrow \frac{AB}{200}=\frac{\sqrt{3}}{2}$$
$$\Rightarrow AB=\left( \frac{200\times \sqrt{3}}{2} \right)m=173.2\text{ }m.$$

Example 5:    The pilot of a helicopter, at an altitude of 1200m finds that the two ships are sailing towards it in the same direction. The angle of depression of the ships as observed from the helicopter are 60º and 45º respectively. Find the distance between the two ships.
Sol.    Let B the position of the helicopter and let C, D be the ships. Let AB be the vertical height.

Then, AB = 1200 m,
∠ACB = 60º and ∠ADB = 45º.
$$\text{Then, }\!\!~\!\!\text{ }\frac{AD}{AB}=\text{cot 45 }\!\!{}^\text{o}\!\!\text{ = 1}$$
$$\Rightarrow \frac{AD}{1200}=1\Rightarrow \text{AD = 1200 m}$$
$$\text{And, }\!\!~\!\!\text{ }\frac{AC}{AB}=\text{cot 60 }\!\!{}^\text{o}\!\!\text{ }=\frac{1}{\sqrt{3}}\text{ }$$
$$\Rightarrow \frac{AC}{1200}=\frac{1}{\sqrt{3}}\text{ }$$
$$\Rightarrow AC=\frac{1200}{\sqrt{3}}=400\sqrt{3}\text{ m}\text{.}$$

Example 6:    A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30º and that of the top of the flagstaff is 45º. Find the height of the tower.
Sol.    Let AB be the tower and BC be the flagstaff.

Then, BC = 7 m. Let AB = h.
Let O be the point of observation.
Then, ∠AOB = 30º and ∠AOC = 45º.
$$\text{Now, }\frac{OA}{AC}=\text{cot 45 }\!\!{}^\text{o}\!\!\text{ }=1$$
⇒  OA = AC = h + 7.
$$\text{And, }\frac{OA}{AB}=\text{cot }30{}^\text{o}=\sqrt{3}$$
$$\Rightarrow \frac{OA}{h}=\sqrt{3}\Rightarrow OA=h\sqrt{3}$$
∴   h + 7 = h√3
$$\Rightarrow \frac{7}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{7(\sqrt{3}+1)}{2}=9.562\text{ }m$$

Example 7:    From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30º and 45º respectively. Find the height of the tower.
Sol.    Let AB be the building and CD be the tower.

Then, AB = 30 m. Let DC = x.
Draw DE ⊥ AB. Then AE = CD = x.
∴   BE = (30 – x) m.
$$\text{Now, }\frac{AC}{AB}=\text{cot 45 }\!\!{}^\text{o}\!\!\text{ }=1$$
$$\Rightarrow \frac{AC}{30}=1\Rightarrow AC=30\text{ }m$$
∴  DE = AC = 30 m.
$$\frac{BE}{DE}=\text{tan }30{}^\text{o}=\frac{1}{\sqrt{3}}\Rightarrow \frac{BE}{30}=\frac{1}{\sqrt{3}}$$
$$\Rightarrow BE=\frac{30}{\sqrt{3}}$$
$$CD=AE=AB-BE=\left( 30-\frac{30}{\sqrt{3}} \right)$$
$$=30\left( 1-\frac{1}{\sqrt{3}} \right)\,m$$

Example 8:    From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower.
Sol.    Let AB be the cliff and CD be the tower.

Then, AB = 25 m. From B draw BE ⊥ CD.
Let ∠EBD = ∠ACB = α.
$$\text{Now, }\frac{\text{DE}}{\text{BE}}=\text{tan }\alpha \text{ and }\frac{\text{AB}}{\text{AC}}=\text{tan }\alpha \text{ }$$
$$\frac{DE}{BE}=\frac{AB}{AC}\text{ }So,\text{ }DE=AB$$ [ ∵ BE = AC]
∴  CD = CE + DE = AB + AB = 2AB = 50m

Example 9:    The altitude of the sun at any instant is 60º. Find the height of the vertical pole that will cast a shadow of 30 m.
Sol.    Let AB be the pole and AC be its shadow.
Then, θ = 60º and AC = 30 m.

$$\frac{AB}{AC}=\text{tan }60{}^\text{o}$$
$$\Rightarrow \frac{AB}{30}=\sqrt{3}\Rightarrow AB=30\sqrt{3}\,\,m$$

Example 10:    When the sun is 30º above the horizontal, Find the length of shadow cast by a building 50m high.
Sol.    Let AB be the building and AC be its shadow.
Then, AB = 50 m and θ = 30º.

$$\frac{AC}{AB}=\text{cot 3}0{}^\text{o}=\sqrt{3}$$
$$\Rightarrow \frac{AC}{50}=\sqrt{3}$$
⇒ AC = 50√3 cm.

Example 11:    If the elevation of the sun changed from 30º to 60º, then find the difference between the lengths of shadows of a pole 15 m high, made at these two positions.
Sol.    When AB = 15m, θ = 30º,
$$\text{then }\frac{AC}{AB}=\text{tan}30{}^\text{o}$$
$$\Rightarrow AC=\frac{15}{\sqrt{3}}m.$$
When AB = 15m, θ = 60º,
$$\text{then }\frac{AC}{AB}=\text{tan6}0{}^\text{o}$$
⇒  AC = 15√3 m.
∴ Diff. in lengths of shadows
$$=\left( 15\sqrt{3}-\frac{15}{\sqrt{3}} \right)$$
$$=\frac{30}{\sqrt{3}}=10\sqrt{3}\,\,m$$

Example 12:    The heights of two poles are 80 m and 62.5 m. If the line joining their tops makes an angle of 45º with the horizontal, then find the distance between the poles.
Sol.    Let AB and CD be the poles such that

AB = 80 m and CD = 62.5 m.
Draw DE ⊥ AB. Then, ∠EDB = 45º
Now, BE = AB – AE = AB – CD = 17.5
$$\frac{DE}{BE}=\text{cot }45{}^\text{o}=1$$
⇒ DE = BE = 17.5 m.

Example 13:    If the angle of elevation of cloud from a point 200 m above a lake is 30º and the angle of depression of its reflection in the lake is 60º, then find the height of the cloud above the lake.
Sol.    Let C be the cloud and C’ be its reflection in the lake.
Let CS = C’S = x.

$$\text{Now, }\frac{BC}{AB}=\text{tan }30{}^\text{o}=\frac{1}{\sqrt{3}}$$
$$\Rightarrow x-200=\frac{AB}{\sqrt{3}}$$
$$\text{Also, }\frac{BC’}{AB}=\text{tan }60{}^\text{o}=\sqrt{3}$$
$$\Rightarrow ~~x+200=(AB)\sqrt{3}.$$
$$\sqrt{3}(x-200)=\frac{x+200}{\sqrt{3}}\text{ or }x=\text{400}.$$
∴  CS = 400 m.

Example 14:    A balloon of radius γ makes an angle α at the eye of an observer and the angle of elevation of its centre is β. Then find the height of its centre from the ground level.
Sol.    Let C be the centre of the balloon and O be the position of the observer at the horizontal line OX. Let OA and OB be the tangents to the balloon so that ∠AOB = α, ∠XOC = β  and
CA = CB = γ.

Clearly, right angled triangles OAC and OBC are congruent.
∠AOC = ∠BOC = α/2.
Let CN ⊥ OX.
$$\text{Now, }\frac{OC}{CA}=\text{cosec }\!\!~\!\!\text{ }\frac{\alpha }{2}$$
$$\Rightarrow OC=\gamma \text{ cosec }\!\!~\!\!\text{ }\frac{\alpha }{2}\text{ }…\text{(i)}$$
$$\text{Also, }\frac{CN}{OC}=\sin \beta$$
$$\Rightarrow CN=OC\text{ sin }\beta =\gamma \text{ cosec }\!\!~\!\!\text{ }\frac{\alpha }{2}\sin \beta \text{ }\left[ \text{Using}\left( \text{i} \right) \right]$$

Example 15:    The banks of a river are parallel.  A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45º and reaches the opposite bank at a point 20 m from the point opposite to the starting point. Find the breadth of the river.
Sol.    Let A be the starting point and B, the end point of the swimmer.  Then AB = 20 m and ∠BAC = 45º.

$$\text{Now, }\frac{BC}{AB}=\text{sin}45{}^\text{o}=\frac{1}{\sqrt{2}}$$
$$\Rightarrow \frac{BC}{20}=\frac{1}{\sqrt{2}}$$
$$\Rightarrow BC=\frac{20\times \sqrt{2}}{2}=14.14\text{ }m.$$

Example 16:    A man on a cliff observes a fishing trawler at an angle of depression of 30º which is approaching the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle of depression of the trawler is found to be 60º. Caliculate the time taken by the trawler to reach the shore.
Sol.    Let AB be the cliff and C and D be the two positions of the fishing trawler.
Then, ∠ACB = 30º and ∠ADB = 60º

Let AB = h.
$$\text{Now, }\frac{AD}{AB}=\text{cot }60{}^\text{o}=~\frac{1}{\sqrt{3}}$$
$$\Rightarrow AD=\frac{h}{\sqrt{3}}$$
$$\text{And, }\frac{AC}{AB}=\text{cot 3}0{}^\text{o}=~\sqrt{3}$$
⇒ AC = √3 h
$$CD=ACAD=\left( \sqrt{3}\,h-\frac{h}{\sqrt{3}} \right)=\frac{2h}{\sqrt{3}}$$
Let u m/min be the uniform speed of the trawler.
Distance covered in 6 min = 6u metres.
$$CD=6u~\Rightarrow \frac{2h}{\sqrt{3}}=6u\Rightarrow h=3\sqrt{3}\,\,u$$
$$Now,\text{ }AD=\frac{h}{\sqrt{3}}=\frac{3\sqrt{3}\,\,u}{\sqrt{3}}=3u$$
Time taken by trawler to reach A
$$=\frac{dis\tan ce\,AD}{speed}\Rightarrow A=\frac{3u}{u}=3\min$$

Example 17:    A boat is being rowed away from a cliff 150m high. At the top of the cliff the angle of depression of the boat changes from 60º to 45º in 2 minutes. Find the speed of the boat.
Sol.    Let AB be the cliff and C and D be the two positions of the ship. Then, AB = 150 m,
∠ACB = 60º and ∠ADB = 45º.

$$Now,\frac{AD}{AB}=\text{cot }45{}^\text{o}=1$$
$$\Rightarrow \frac{AD}{150}=1\Rightarrow AD\text{ }=\text{ }150\text{ }m.$$
$$\frac{AC}{AB}=\text{cot 60}{}^\text{o}=\frac{1}{\sqrt{3}}\Rightarrow \frac{AC}{150}=\frac{1}{\sqrt{3}}$$
$$\Rightarrow AC=\frac{150}{\sqrt{3}}=50\sqrt{3}=\text{ }86.6\text{ }m$$
∴ CD = AD – AC = (150 – 86.6) m = 63.4 m
Thus, distance covered in 2 min. = 63.4 m
∴ Speed of the boat
$$=\left( \frac{63.4}{2}\times \frac{60}{1000} \right)\,\,km/hr.=\text{ }1.9\text{ }km/hr.$$

Example 18:    A tower is 100√3 metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.
Sol.    Let AB be the tower of height 100√3 metres, and let C be a point at a distance of 100 metres from the foot of the tower.

Let θ be the angle of elevation of the top of the tower from point C.
In ∆CAB, we have
$$\tan \theta =\frac{AB}{AC} \\$$
$$\Rightarrow \tan \theta =\frac{100\sqrt{3}}{100}=\sqrt{3}$$
⇒ θ = 60º
Hence, the angle of elevation of the top of the tower from a point 100 metres away from its foot is 60º.

Example 19:    From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30º. The angle of elevation of the top of a water tank (on the top of the tower) is 45º. Find the (i) height of the tower (ii) the depth of the tank.
Sol.    Let BC be the tower of height h metre and CD be the water tank of height h1 metre.
Let A be a point on the ground at a distance of 40 m away from the foot B of the tower.

In ∆ABD, we have tan 45º = $$\frac { BD }{ AB }$$
$$\Rightarrow 1=\frac{h+{{h}_{1}}}{40}\Rightarrow ~~h+{{h}_{1}}=40\text{ }m~\text{ }\text{……}\left( \text{i} \right)$$
In ∆ABC, we have
$$\text{tan }30{}^\text{o}\text{ }=\frac{BC}{AB}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40}$$
$$\Rightarrow h=\frac{40}{\sqrt{3}}=\frac{40\sqrt{3}}{3}=23.1\text{ m}$$
Substituting the value of h in (i), we get
23.1 + h1 = 40
⇒ h1 = (40 – 23.1)m = 16.9 m

Example 20:    Two stations due south of a leaning tower which leans towards the north are at distance a and b from its foot. If α, β be the elevations of the top of the tower from these stations, prove that its inclination θ to the horizontal is given by
$$\text{cot }\theta =\frac{b\,\,\cot \,\,\alpha -a\,\,\cot \,\,\beta }{b-a}$$
Sol.    Let AB be the leaning tower and let C and D be two given stations at distances a and b respectively from the foot A of the tower.

Let AE = x and BE = h
In ∆ABE, we have
$$\tan \theta =\frac{BE}{AE}\Rightarrow \tan \theta =\frac{h}{x}$$
⇒ x = h cot θ               ….(i)
In ∆CBE, we have
$$\tan \alpha =\frac{BE}{CE}$$
$$\Rightarrow \tan \alpha =\frac{h}{a+x}$$
⇒ a + x = h cot α
⇒ x = h cot α – a      ….(ii)
In ∆DBE, we have
$$\tan \beta =\frac{BE}{DE}$$
$$\Rightarrow \tan \beta =\frac{h}{b+x}$$
⇒ b + x = h cot β
⇒ x = h cot β– b       ….(iii)
From equations (i) and (ii), we have
h cot θ= h cot α– a
⇒ h (cot α– cot θ) = a
$$\Rightarrow h=\frac{a}{\cot \alpha -\cot \theta }$$
From equation (i) and (iii), we get
h cot θ= h cot β– b
⇒ h (cot β– cot θ) = b
$$\Rightarrow h=\frac{b}{\cot \beta -\cot \theta }$$
Equating the values of h from equations (iv) and (v), we get
$$\frac{a}{\cot \alpha -\cot \theta }=\frac{b}{\cot \beta -\cot \theta }$$
⇒ a(cot β– cot θ) = b(cot α– cot θ)
⇒ (b – a) cot θ= b cot α– a cot β
$$\cot \theta =\frac{b\,\,\cot \alpha -a\cot \beta }{b-a}$$

Example 21:    If the angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the height of the cloud is $$\frac{h(\tan \alpha +\tan \beta \,)}{\tan \beta -\tan \alpha }$$.
Sol.    Let AB be the surface of the lake and let P be a point of observation such that AP = h metres. Let C be the position of the cloud and C´ be its reflection in the lake. Then, CB = C´B. Let PM be perpendicular from P on CB. Then,
∠CPM = α and ∠MPC´= β Let CM = x.
Then, CB = CM + MB = CM + PA = x + h.

In ∆CPM, we have
$$\tan \alpha =\frac{CM}{PM}$$
$$\Rightarrow \tan \alpha =\frac{x}{AB}$$
⇒ AB = x cot α            ….(i)
In ∆PMC´, we have
$$\tan \beta =\frac{C\acute{\ }M}{PM}$$
$$\Rightarrow \tan \beta =\frac{x+2h}{AB}$$
[∵ C´M = C´B + BM = x + h + h]
⇒ AB = (x + 2h) cot β          ….(ii)
From (i) and (ii), we have
x cot α= (x + 2h) cot β
⇒ x(cot α– cot β) = 2h cot β
$$\Rightarrow x\left( \frac{1}{\tan \alpha }-\frac{1}{\tan \beta } \right)=\frac{2h}{\tan \beta }$$
$$\Rightarrow x\left( \frac{\tan \beta -\tan \alpha }{\tan \alpha \tan \beta } \right)=\frac{2h}{\tan \beta }$$
$$\Rightarrow x=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }$$
Hence,
Height of the cloud = x + h
$$=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }+h$$
$$=\frac{2h\tan \alpha +h\tan \beta -h\tan \alpha }{\tan \beta -\tan \alpha }$$
$$=\frac{h(\tan \alpha +\tan \beta \,)}{\tan \beta -\tan \alpha }$$

Example 22:    There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 30º and 45º, find the height of the tree.
Sol.     Let OA be the tree of height h metre.
In triangles POA and QOA, we have
$$\text{tan }30{}^\text{o}=\frac{OA}{OP}\text{ and tan }45{}^\text{o}=\frac{OA}{OQ}$$

$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{OP}\text{ and }1=\frac{h}{OQ}$$
⇒ OP = √3 h and OQ = h
⇒ OP + OQ = √3 h + h ⇒ PQ = (√3+ 1)h
⇒ 100 = (√3 + 1)h [∵ PQ = 100 m]
$$\Rightarrow h=\frac{100}{\sqrt{3}+1}\Rightarrow h=\frac{100(\sqrt{3}-1)}{2}\text{ m}$$
⇒ h = 50(1.732 – 1) m = 36.6 m
Hence, the height of the tree is 36.6 m

Example 23:    The angle of elevation of a cliff from a fixed point is θ. After going up a distance of k metres towards the top of cliff at an angle of φ, it is found that the angle of elevation is α. Show that the height of the cliff is metres
Sol.     Let AB be the cliff and O be the fixed point such that the angle of elevation of the cliff from O is θ i.e. ∠AOB = θ. Let ∠AOC = φ and
OC = k metres. From C draw CD and CE perpendiculars on AB and OA respectively.
Then, ∠DCB = α.
Let h be the height of the cliff AB.

In ∆OCE, we have
$$\sin \phi =\frac{CE}{OC}$$
$$\Rightarrow \sin \phi =\frac{CE}{k}$$
⇒ CE = k sin φ       …(i)     [∵ CE = AD]
⇒ AD = k sin φ
$$\text{And, cos}\phi =\frac{OE}{OC}$$
$$\Rightarrow \text{cos}\phi =\frac{OE}{k}$$
⇒ OE = k cos φ          ….(ii)
In ∆ OAB, we have
$$\text{tan }\theta =\frac{AB}{OA}$$
$$\Rightarrow \text{tan }\theta =\frac{h}{OA}$$
⇒ OA = h cot θ          ….(iii)
CD = EA = OA – OE
= h cot θ – k cos φ        …..(iv)      [Using eqs.(ii) and (iii)]
and, BD = AB – AD = AB – CE
= (h – k sin φ)         ….(v)       [Using equation (i)]
In ∆BCD, we have
$$\text{tan}\alpha =\frac{BD}{CD}\Rightarrow \text{ tan}\alpha =\frac{h-k\sin \varphi }{h\cot \theta -k\cos \varphi }$$ [Using equations (iv) and (v)]
$$\Rightarrow \frac{1}{\cot \alpha }=\frac{h-k\sin \varphi }{h\cot \theta -k\cos \varphi}$$
⇒ h cot α – k sin φ cot α = h cot θ – k cos φ
⇒ h(cot θ – cot α) = k(cos φ – sin φcot α)
$$\Rightarrow h=\frac{k(\cos \varphi -\sin \varphi \cot \alpha )}{\cot \theta -\cot \alpha }$$

Example 24:    At the foot of a mountain the elevation of its summit is 45º; after ascending 1000 m towards the mountain up a slope of 30º inclination is found to be 60º. Find the height of the mountain.
Sol.    Let F be the foot and S be the summit of the mountain FOS. Then, ∠OFS = 45º and therefore ∠OSF = 45º. Consequently,
OF = OS = h km(say).
Let FP = 1000 m = 1 km be the slope so that
∠OFP = 30º. Draw PM ⊥OS and PL ⊥OF.
Join PS. It is given that ∠MPS = 60º.
In ∆FPL, We have

$$\text{sin }30{}^\text{o}=\frac{PL}{PF}$$
$$\Rightarrow PL=PF\text{ sin }30{}^\text{o}=\left( 1\times \frac{1}{2} \right)=\frac{1}{2}km.$$
$$OM=PL=\frac{1}{2}km$$
$$\Rightarrow MS=OS-OM=~\left( h-\frac{1}{2} \right)km~~~~\ldots .\left( i \right)$$
$$\text{Also, cos }30{}^\text{o}=\frac{FL}{PF}$$
$$\Rightarrow FL=PF\text{ cos }30{}^\text{o}=\left( 1\times \frac{\sqrt{3}}{2} \right)=\frac{\sqrt{3}}{2}km$$
Now, h = OS = OF = OL + LF
$$\Rightarrow h=OL+\frac{\sqrt{3}}{2}$$
$$\Rightarrow OL=\left( h-\frac{\sqrt{3}}{2} \right)km$$
$$\Rightarrow PM=\left( h-\frac{\sqrt{3}}{2} \right)km$$
In ∆PSM, we have
$$\text{tan }60{}^\text{o}=\frac{SM}{PM}$$
⇒ SM = PM. tan 60º …..(ii)
$$\Rightarrow \left( h-\frac{1}{2} \right)=\left( h-\frac{\sqrt{3}}{2} \right)\sqrt{3}$$    [Using equations (i) and (ii)]
$$\Rightarrow h-\frac{1}{2}=h\sqrt{3}-\frac{3}{2}$$
⇒ h(√3 – 1) = 1
$$\Rightarrow h=\frac{1}{\sqrt{3}-1}$$
$$\Rightarrow h=\frac{\sqrt{3}+1}{(\sqrt{3}-1)\,(\sqrt{3}+1)}=\frac{\sqrt{3}+1}{2}$$
$$=\frac{2.732}{2}=1.336\text{ }km$$
Hence, the height of the mountain is 1.366 km.

Example 25:    The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of the tower is β. If AB = d, show that the height of the tower is  $$\frac{d}{\sqrt{{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta }}$$
Sol.   Let OP be the tower and let A and B be two points due south and east respectively of the tower such that ∠OAP = αand ∠OBP = β.
Let OP = h. In ∆OAP, we have

$$\tan \alpha =\frac{h}{OA}$$
⇒ OA = h cot α             ….(i)
In ∆OBP, we have
$$\tan \beta =\frac{h}{OB}$$
⇒ OB = h cot β.            ….(ii)
Since OAB is a right angled triangle. Therefore,
AB2 = OA2 + OB2
⇒ d2 = h2 cot2 α + h2 cot2 β
$$\Rightarrow h=\frac{d}{\sqrt{{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta }}$$    [Using (i) and (ii)]

Example 26:     The elevation of a tower at a station A due north of it is α and at a station B due west of A is β. Prove that the height of the tower is   $$\frac{AB\sin \alpha \sin \beta }{\sqrt{{{\sin }^{2}}\alpha -{{\sin }^{2}}\beta }}$$
Sol.   Let OP be the tower and let A be a point due north of the tower OP and let B be the point due west of A. Such that ∠OAP = and ∠OBP = Let h be the height of the tower.
In right angled triangles OAP and OBP, we have

$$\tan \alpha =\frac{h}{OA}\text{ and }\tan \beta =\frac{h}{OB}$$
⇒ OA = h cot α and OB = h cot β.
In ∆OAB, we have
OB2 = OA2 + AB2
⇒ AB2 = OB2 – OA2
⇒ AB2 = h2 cot2 β– h2 cot2 α
⇒ AB2 = h2 [cot2 β– cot2 α]
⇒ AB2 = h2[(cosec2 β– 1) – (cosec2 α– 1)]
⇒ AB2 = h2(cosec2 β– cosec2 α)
$$\Rightarrow A{{B}^{2}}={{h}^{2}}\left( \frac{{{\sin }^{2}}\alpha -{{\sin }^{2}}\beta }{{{\sin }^{2}}\alpha {{\sin }^{2}}\beta } \right)$$
$$\Rightarrow h=\frac{AB\sin \alpha \sin \beta }{\sqrt{{{\sin }^{2}}\alpha -{{\sin }^{2}}\beta }}$$

Example 27:     An aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60º and 45º respectively. Find the vertical distance between the aeroplanes at that instant.
Sol.    Let P and Q be the positions of two aeroplanes when Q is vertically below P and OP = 4000 m. Let the angles of elevation of P and Q at a point A on the ground be 60º and 45º respectively.

$$\text{tan }60{}^\text{o}=\frac{OP}{OA}\text{ and tan 45}{}^\text{o}=\frac{OQ}{OA}$$
$$\Rightarrow \sqrt{3}=\frac{4000}{OA}\text{ and 1=}\frac{OQ}{OA}$$
$$\Rightarrow OA=\frac{4000}{\sqrt{3}}\text{ and }OQ=OA$$
$$\Rightarrow OQ=\frac{4000}{\sqrt{3}}m$$
In triangles AOP and AOQ, we have
∴ Vertical distance between the aeroplanes
= PQ = OP – OQ
$$=\left( 4000-\frac{4000}{\sqrt{3}} \right)=4000\frac{(\sqrt{3}-1)}{\sqrt{3}}m$$
= 1690.53 m

## Trigonometry: Solving Word Problems

Trigonometry can be used on a daily basis in the workplace. Since trigonometry means “triangle measure”, any profession that deals with measurement deals with trigonometry as well. Carpenters, construction workers and engineers, for example, must possess a thorough understanding of trigonometry.

In word problems, the formulas remain the same:

### Word problems introduce two new vocabulary terms:

Angle of Elevation:
In this diagram, xº marks theangle of elevation of the top of the tree as seen from a point on the ground.

The angle of elevation is always measured from the ground up. Think of it like an elevator that only goes up. It is always INSIDE the triangle.

You can think of the angle of elevation in relation to the movement of your eyes. You are looking straight ahead and you must raise (elevate) your eyes to see the top of the tree.

Angle of Depression:
In this diagram, xº marks the angle of depression of the boat at sea from the top of the lighthouse.

The angle of depression is always OUTSIDE the triangle. It is never inside the triangle.

You can think of the angle of depression in relation to the movement of your eyes. You are standing at the top of the lighthouse and you are looking straight ahead. You must lower (depress) your eyes to see the boat in the water.

Notice how the horizontal line in the angle of depression diagram is PARALLEL to the ground level. The fact that horizontal lines are always parallel guarantees that the alternate interior angles are equal in measure. In the diagram, the angle marked xº is equal in measure to m∠BAC. Simply stated, this means that …
the angle of elevation = the angle of depression.

So what do we do with this angle of depression that is OUTSIDE of our triangle?
There are two possible ways to use our angle of depression to obtain an angle INSIDE the triangle.
Option 1: find the angle inside the triangle that is adjacent (next door) to the angle of depression.

This adjacent angle will always be the complement of the angle of depression, since the horizontal line and the vertical line are perpendicular (90º). In the diagram below, the adjacent angle is 52º.

Option 2:

utilize the fact that the angle of depression = the angle of elevation and label ∠BAC as 38º inside the triangle.

Notice that both options, the answer is the same.

Example 1:
A nursery plants a new tree and attaches a guy wire to help support the tree while its roots take hold. An eight foot wire is attached to the tree and to a stake in the ground. From the stake in the ground the angle of elevation of the connection with the tree is 42º. Find to the nearest tenth of a foot, the height of the connection point on the tree.

Solution:

• A “guy” wire is a support wire used to hold a newly planted tree in place, preventing it from bending or up-rooting during high winds.
• The “angle of elevation” is from the ground up.
• It is assumed that the tree is vertical, making it perpendicular with the ground.
• This problem deals with “opposite” and “hypotenuse” making it a sine problem.

Example 2: From the top of a fire tower, a forest ranger sees his partner on the ground at an angle of depression of 40º. If the tower is 45 feet in height, how far is the partner from the base of the tower, to the nearest tenth of a foot?

Solution:

• Remember that the “angle of depression” is from a horizontal line of sight downward.
• It is assumed that the tower is vertical, making it perpendicular with the ground.
• This solution will use alternate interior angles from the parallel horizontal lines, so place 40º inside the triangle by the partner (bottom right).
• This solution deals with “opposite” and “adjacent” making it a tangent problem.

Example 3: Find the shadow cast by a 10 foot lamp post when the angle of elevation of the sun is 58º. Find the length to the nearest tenth of a foot.

Solution:

• Remember that the “angle of elevation” is from the horizontal ground line upward.
• It is assumed that the lamp post is vertical, making it perpendicular with the ground.
• Shadows are on the ground! If you place the “shadow” on the hypotenuse you have created an apparition ( a “ghost”), not a shadow!
• This solution deals with “opposite” and “adjacent” making it a tangent problem.

Example 4: A ladder leans against a brick wall. The foot of the ladder is 6 feet from the wall. The ladder reaches a height of 15 feet on the wall. Find to the nearest degree, the angle the ladder makes with the wall.

Solution:

• In this problem place xº where the ladder meets the wall. Do not assume that the angle will always be at the ground level.
• It is assumed that the wall is vertical, perpendicular with the ground.
• The foot of the ladder is the bottom of the ladder, where it hits the ground.
• This solution deals with “opposite” and “adjacent” making it a tangent problem.

Example 5: A radio station tower was built in two sections. From a point 87 feet from the base of the tower, the angle of elevation of the top of the first section is 25º, and the angle of elevation of the top of the second section is 40º. To the nearest foot, what is the height of the top section of the tower?

Solution:

• Think of this problem as working with two separate triangles:
(1) the larger triangle with the 40º angle and a vertical side that represents the ENTIRE height, b, of the tower, and
(2) the smaller triangle with the 25º angle and a vertical side, a, that represents the height of the first (bottom) section of the tower.
• Solve for the vertical heights (b and a) in the two separate triangles.
• The needed height, x, of the second (top) section of the tower will be the difference between the ENTIRE height, b, and the height of the first (bottom) section, a. You will need to subtract.
• In both triangles, the solution deals with “opposite” and “adjacent” making it a tangent problem.
• Difference (b – a) = 73.00166791 – 40.56876626 = 32.43290165 ≈ 32 feet

## Trigonometry: Solving for an Angle

### Using Trigonometric Functions to Find a Missing Angle

The initial set up for solving these problems will be the same as that for finding a missing side.

To finish the problem, however, it will be necessary to use a calculator function referred to as an “inverse function” to find the actual number of degrees in the angle.

The inverse functions, on the graphing calculator, for each of the three trigonometric functions are found directly above the buttons for sine, cosine and tangent. They appear as sin-1, cos-1 and tan-1.
Note: sin-1(x) is read “the angle whose sine is x”.

You will discover, in later courses, that there are actually many angles whose sine is x, but in this course, we are looking for the simplest, most basic angle that has a sine x.

You can think of the inverse functions as “undoing” the trigonometric functions, leaving us with just the angle.
sin-1(sin x) = x
As your study of trigonometric inverse functions continues, you will see that sin-1(x), cos-1(x) and tan-1(x) may also be written as arcsin(x), arccos(x), and arctan(x), which are read “the arc whose sine is x”, and so on.

Set Up the Diagram:
Find x, to the nearest degree.

Notice that in this problem, the x is INSIDE the triangle representing the angle. The a is alone, so this problem deals with o and h, which is sine.

Set Up the Formula:

Now, divide 30 by 40 to change the fraction to a decimal.
sin x = 0.75
The goal now is to find an angle whose sine is 0.75. To do this, use the sin-1 function on your calculator!
On the graphing calculator: activate sin-1 (above the sin key) and then enter 0.75.
On the scientific calculator: enter 0.75 and then activate the sin-1 above the sin key.
Be sure you are in DEGREE MODE.