Some Applications of Trigonometry

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Answer

From figure,

AB is the pole.

In △BAC,

sin 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{2} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{AB}{20} \\[1em] \Rightarrow AB = \dfrac{1}{2} \times 20 = 10 \text{ m}.$

Hence, height of pole = 10 meters.

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer

From figure,

Let AB be the left over tree and BC be the broken part of tree.

C is the top of tree and A is the foot of the tree.

In △ABC,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{8} \\[1em] \Rightarrow AB = \dfrac{8}{\sqrt{3}} \text{ m}.$

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ BC² = AB² + AC²

$\Rightarrow BC^2 = \Big(\dfrac{8}{\sqrt{3}}\Big)^2 + 8^2 \\[1em] \Rightarrow BC^2 = \dfrac{64}{3} + 64 \\[1em] \Rightarrow BC^2 = \dfrac{64 + 192}{3} \\[1em] \Rightarrow BC^2 = \dfrac{256}{3} \\[1em] \Rightarrow BC = \sqrt{\dfrac{256}{3}} \\[1em] \Rightarrow BC = \dfrac{16}{\sqrt{3}} \text{ m}.$

Height of tree = AB + BC

$= \dfrac{8}{\sqrt{3}} + \dfrac{16}{\sqrt{3}} \\[1em] = \dfrac{24}{\sqrt{3}} \\[1em] = \dfrac{24}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] = \dfrac{24\sqrt{3}}{3} \\[1em] = 8\sqrt{3} \text{ m}.$

Hence, height of tree = $8\sqrt{3}$ m.

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer

Let AC be the slide for children below 5 years and DF be the slide for elder children.

In △ABC,

sin 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{2} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{1.5}{AC} \\[1em] \Rightarrow AC = 1.5 \times 2 = 3 \text{ m}.$

In △DEF,

sin 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{DE}{DF} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{3}{DF} \\[1em] \Rightarrow DF = \dfrac{3 \times 2}{\sqrt{3}} = 2\sqrt{3} \text{ m}.$

Hence, for children below 5 years, length of slide = 3 m and for elder children, length of slide = $2\sqrt{3}$ m.

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer

From figure,

Let AB be the tower.

In △ABC,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{30} \\[1em] \Rightarrow AB = \dfrac{30}{\sqrt{3}} \\[1em] \Rightarrow AB = 10\sqrt{3} \text{ m}.$

Hence, height of tower = $10\sqrt{3}$ m.

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer

Let A be the point where kite is present, AC is the string and C is the point where string is tied on the ground.

In △ABC,

sin 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{60}{AC} \\[1em] \Rightarrow AC = \dfrac{60 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{120}{\sqrt{3}}$

Multiplying numerator and denominator by $\sqrt{3}$,

$\Rightarrow AC = \dfrac{120}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{120\sqrt{3}}{3} \\[1em] \Rightarrow AC = 40\sqrt{3} \text{ m}.$

Hence, length of string = $40\sqrt{3}$ m.

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Let AB be the initial position of boy and CD be the tower.

From figure,

CG = CD - GD = 30 - 1.5 = 28.5 m

In △AGC,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CG}{AG} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{28.5}{AG} \\[1em] \Rightarrow AG = 28.5 \times \sqrt{3} \\[1em] \Rightarrow AG = 28.5\sqrt{3} \text{ m}.$

In △EGC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{CG}{EG} \\[1em] \Rightarrow \sqrt{3} = \dfrac{28.5}{EG} \\[1em] \Rightarrow EG = \dfrac{28.5}{\sqrt{3}} \text{ m}.$

From figure,

$\Rightarrow AE = AG - EG \\[1em] = 28.5\sqrt{3} - \dfrac{28.5}{\sqrt{3}} \\[1em] = \dfrac{28.5 \times 3 - 28.5}{\sqrt{3}} \\[1em] = \dfrac{85.5 - 28.5}{\sqrt{3}} \\[1em] = \dfrac{57}{\sqrt{3}} \\[1em] = \dfrac{57}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] = \dfrac{57\sqrt{3}}{3} \\[1em] = 19\sqrt{3} \text{ m}.$

Hence, the boy walked $19\sqrt{3}$ m towards the building.

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer

Let BC be the building and CD be the transmission tower.

In △ABC,

tan 45° = $\dfrac{\text{Side opposite to angle 45°}}{\text{Side adjacent to angle 45°}}$

Substituting values we get :

$\Rightarrow 1 = \dfrac{BC}{AB} \\[1em] \Rightarrow AB = BC = 20 \text{ meters}.$

In △ABD,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{BD}{AB} \\[1em] \Rightarrow BD = AB\sqrt{3} = 20\sqrt{3} \text{ meters}.$

From figure,

CD = BD - BC = $20\sqrt{3} - 20 = 20(\sqrt{3} - 1)$ meters.

Hence, height of tower = $20(\sqrt{3} - 1)$ meters.

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer

Let BC be the pedestal and CD the statue.

In △ABC,

tan 45° = $\dfrac{\text{Side opposite to angle 45°}}{\text{Side adjacent to angle 45°}}$

Substituting values we get :

$\Rightarrow 1 = \dfrac{BC}{AB} \\[1em] \Rightarrow AB = BC = x \text{ (let)}.$

In △ABD,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{BD}{AB} \\[1em] \Rightarrow BD = AB\sqrt{3} = x\sqrt{3} \text{ meters}.$

From figure,

$\Rightarrow CD = BD - BC \\[1em] \Rightarrow 1.6 = x\sqrt{3} - x \\[1em] \Rightarrow x(\sqrt{3} - 1) = 1.6 \\[1em] \Rightarrow x = \dfrac{1.6}{\sqrt{3} - 1}$

Rationalising x, we get :

$\Rightarrow x = \dfrac{1.6}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow x = \dfrac{1.6(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow x = \dfrac{1.6(\sqrt{3} + 1)}{2} \\[1em] \Rightarrow x = 0.8(\sqrt{3} + 1) \text{ meters}.$

Hence, height of pedestal = $0.8(\sqrt{3} + 1)$ meters.

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer

Let AB be the building and CD be the tower.

In △BCD,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{CD}{BD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{50}{BD} \\[1em] \Rightarrow BD = \dfrac{50}{\sqrt{3}} \text{ meters}.$

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{\dfrac{50}{\sqrt{3}}} \\[1em] \Rightarrow AB = \dfrac{50}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{50}{3} = 16\dfrac{2}{3} \text{ meters}.$

Hence, height of building = $16\dfrac{2}{3}$ meters.

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer

Let AB be the first pole and CD be the second pole of each height h meters and E be the point between the road.

Let BE = x meters and ED = (80 - x) meters.

In △ABE,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{x} \\[1em] \Rightarrow h = x\sqrt{3} \text{ .......(1)}$

In △CED,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{ED} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{80 - x} \\[1em] \Rightarrow h = \dfrac{80 - x}{\sqrt{3}} \text{ .......(2)}$

From (1) and (2), we get :

$\Rightarrow x\sqrt{3} = \dfrac{80 - x}{\sqrt{3}} \\[1em] \Rightarrow 3x = 80 - x \\[1em] \Rightarrow 3x + x = 80 \\[1em] \Rightarrow 4x = 80 \\[1em] \Rightarrow x = \dfrac{80}{4} = 20 \text{ m}.$

Substituting value of x in equation (1), we get :

⇒ h = $20\sqrt{3}$ meters.

⇒ 80 - x = 80 - 20 = 60 meters.

Hence, height of poles = $20\sqrt{3}$ meter and distance of point from first tower = 20 m and from second tower = 60 m.

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Answer

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = AB\sqrt{3} \text{ m}.$

In △ABC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \text{ m}.$

From figure,

CD = BD - BC

$\Rightarrow 20 = AB\sqrt{3} - \dfrac{AB}{\sqrt{3}} \\[1em] \Rightarrow 20 = \dfrac{3AB - AB}{\sqrt{3}} \\[1em] \Rightarrow 20 = \dfrac{2AB}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{20\sqrt{3}}{2} \\[1em] \Rightarrow AB = 10\sqrt{3} \text{ m}.$

Width of canal (BC) = $\dfrac{AB}{\sqrt{3}} = \dfrac{10\sqrt{3}}{\sqrt{3}}$ = 10 m.

Hence, height of tower = $10\sqrt{3}$ m and width of canal = 10 m.

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer

Let AB be the building and CD be the cable tower.

Given,

Angle of depression of foot of tower from top of building is 45°.

∴ ∠EAC = 45°

From figure,

∠ACB = ∠EAC = 45° [Alternate angles are equal].

In △ABC,

tan 45° = $\dfrac{\text{Side opposite to angle 45°}}{\text{Side adjacent to angle 45°}}$

Substituting values we get :

$\Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = AB = 7 \text{ m}.$

From figure,

AE = BC = 7 m.

In △ADE,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{7} \\[1em] \Rightarrow DE = 7\sqrt{3} \text{ m}.$

From figure,

CD = CE + DE = AB + DE = $7 + 7\sqrt{3} = 7(1 + \sqrt{3})$ m.

Hence, height of tower = $7(1 + \sqrt{3})$ meters.

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer

Let AB be the lighthouse and C and D be the position of ships.

We know that,

Alternate angles are equal.

From figure,

⇒ ∠ACB = ∠EAC = 45°

⇒ ∠ADB = ∠EAD = 30°

In △ABC,

tan 45° = $\dfrac{\text{Side opposite to angle 45°}}{\text{Side adjacent to angle 45°}}$

Substituting values we get :

$\Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = AB = 75 \text{ m}.$

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = AB\sqrt{3} = 75\sqrt{3} \text{ m.} \\[1em]$

From figure,

CD = BD - BC = $75\sqrt{3} - 75 = 75(\sqrt{3} - 1)$.

Hence, distance between two ships = $75(\sqrt{3} - 1)$ meters.

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Answer

From figure,

FE = FD - ED = FD - AB = 88.2 - 1.2 = 87 m.

GH = FE = 87 m

In △AGH,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{GH}{AH} \\[1em] \Rightarrow AH = \dfrac{GH}{\sqrt{3}} \\[1em] \Rightarrow AH = \dfrac{87}{\sqrt{3}} \text{ m}.$

In △AEF,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{FE}{AE} \\[1em] \Rightarrow AE = FE\sqrt{3} = 87\sqrt{3} \text{ m}.$

From figure,

EH = AE - AH

= $87\sqrt{3} - \dfrac{87}{\sqrt{3}}$

= $\dfrac{87 \times 3 - 87}{\sqrt{3}}$

= $\dfrac{261 - 87}{\sqrt{3}} = \dfrac{174}{\sqrt{3}}$

= $58\sqrt{3}$ m.

Hence, distance travelled by the balloon during the interval = $58\sqrt{3}$ m.

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer

Let AB be the tower, A position of man, D be the initial position of car and C be the position of car after 6 seconds.

We know that,

Alternate angles are equal.

From figure,

⇒ ∠ADB = ∠EAD = 30°

⇒ ∠ACB = ∠EAC = 60°

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = AB\sqrt{3} \text{ m}.$

In △ABC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \text{ m}.$

From figure,

CD = BD - BC

= $AB\sqrt{3} - \dfrac{AB}{\sqrt{3}} = \dfrac{3AB - AB}{\sqrt{3}} = \dfrac{2AB}{\sqrt{3}}$

= $2 \times \dfrac{AB}{\sqrt{3}} = 2 \times BC$.

According to question,

It takes 6 seconds to cover distance CD or 2BC

and the car is moving with uniform speed.

Since, in 6 seconds car covers a distance of 2BC,

So, a distance of BC meters will be covered in $\dfrac{6}{2}$ = 3 seconds.

Hence, car will take 3 seconds to reach the foot of tower.