Some Applications of Trigonometry | Exercise 9.1

Question 15

  1. 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∘30^\circ, 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∘60^\circ. Find the time taken by the car to reach the foot of the tower from this point.
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Solution

Angle of Depression: The angle from the horizontal downward to a point below. Here, from tower top A, the depression angle to the car's initial position C is 30°, and after 6 seconds to position D is 60°.

Alternate Interior Angles: The depression angle at A equals the elevation angle at the car's position (horizontal from A ∥ ground, with the line of sight as transversal). So ∠ACB=30°\angle ACB = 30° and ∠ADB=60°\angle ADB = 60°.

Tangent Ratio: tan⁡θ=Tower heightHorizontal distance\tan\theta = \dfrac{\text{Tower height}}{\text{Horizontal distance}} — used to express CB and DB in terms of height hh.

Car moving toward tower: As the car approaches, the angle of depression increases (30° → 60°). D is closer to the tower than C, so CB=CD+DBCB = CD + DB.

We will use trigonometric ratios to find distances and then relate them using speed and time.

Step 1 — Define variables and set up equations

Let's draw a diagram for clarity. Let ABAB be the tower. Let hh be the height of the tower. Let CC be the initial car position. Let DD be the car's position after 6 seconds. Let BB be the foot of the tower. The angle of depression from AA to CC is 30∘30^\circ. So, ∠ACB=30∘\angle ACB = 30^\circ. The angle of depression from AA to DD is 60∘60^\circ. So, ∠ADB=60∘\angle ADB = 60^\circ. Let DB=xDB = x. Let CD=yCD = y. In right-angled triangle △ABD\triangle ABD: tan⁡60∘=ABDB\tan 60^\circ = \frac{AB}{DB} 3=hx\sqrt{3} = \frac{h}{x} x=h3x = \frac{h}{\sqrt{3}}

In right-angled triangle △ABC\triangle ABC: tan⁡30∘=ABCB\tan 30^\circ = \frac{AB}{CB} 13=hCB\frac{1}{\sqrt{3}} = \frac{h}{CB} CB=h3CB = h\sqrt{3}

Diagram 1

Step 2 — Calculate distance CD

Since D is between C and B on the highway (car has moved closer), the total distance CB = CD + DB, giving CD = CB − DB.

We know CB=CD+DBCB = CD + DB. So, CD=CB−DBCD = CB - DB. Let's substitute the values from Step 1. CD=h3−h3CD = h\sqrt{3} - \frac{h}{\sqrt{3}} CD=h(3−13)CD = h \left( \frac{3 - 1}{\sqrt{3}} \right) CD=h(23)CD = h \left( \frac{2}{\sqrt{3}} \right)

CD=2h3\boxed{CD = \frac{2h}{\sqrt{3}}}

Step 3 — Calculate the car's speed

The car covers distance CDCD in 6 seconds. Let vv be the uniform speed of the car. Speed Formula: Speed=DistanceTime\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}, so Time=DistanceSpeed\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}.

We use the formula: Speed = Distance / Time. v=CD6v = \frac{CD}{6} v=2h36v = \frac{\frac{2h}{\sqrt{3}}}{6} v=2h63v = \frac{2h}{6\sqrt{3}} v=h33v = \frac{h}{3\sqrt{3}}

v=h33 units/second\boxed{v = \frac{h}{3\sqrt{3}} \text{ units/second}}

Step 4 — Find the time from D to B

We need to find the time to cover distance DBDB. Let tt be this time. We know DB=h3DB = \frac{h}{\sqrt{3}} from Step 1. We know v=h33v = \frac{h}{3\sqrt{3}} from Step 3. We use the formula: Time = Distance / Speed. t=DBvt = \frac{DB}{v} t=h3h33t = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}} t=h3×33ht = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} t=3t = 3

3 seconds\boxed{3 \text{ seconds}}

Answer

The time taken by the car to reach the foot of the tower from this point is 3 seconds.

More questions in Exercise 9.1

Q1

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∘30^\circ (see Fig. 9.11).

Q2

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∘30^\circ 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.

Q3

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∘30^\circ 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∘60^\circ to the ground. What should be the length of the slide in each case?

Q4

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∘30^\circ. Find the height of the tower.

Q5

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∘60^\circ. Find the length of the string, assuming that there is no slack in the string.

Q6

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∘30^\circ to 60∘60^\circ as he walks towards the building. Find the distance he walked towards the building.

Q7

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∘45^\circ and 60∘60^\circ respectively. Find the height of the tower.

Q8
  1. 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∘60^\circ and from the same point the angle of elevation of the top of the pedestal is 45∘45^\circ. Find the height of the pedestal.
Q9
  1. The angle of elevation of the top of a building from the foot of the tower is 30∘30^\circ and the angle of elevation of the top of the tower from the foot of the building is 60∘60^\circ. If the tower is 50 m high, find the height of the building.
Q10
  1. 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∘60^\circ and 30∘30^\circ, respectively. Find the height of the poles and the distances of the point from the poles.
Q11
  1. 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∘60^\circ. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30∘30^\circ (see Fig. 9.12). Find the height of the tower and the width of the canal.
Q12
  1. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60∘60^\circ and the angle of depression of its foot is 45∘45^\circ. Determine the height of the tower.
Q13
  1. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30∘30^\circ and 45∘45^\circ. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Q14
  1. 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∘60^\circ. After some time, the angle of elevation reduces to 30∘30^\circ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
Q15
  1. 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∘30^\circ, 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∘60^\circ. Find the time taken by the car to reach the foot of the tower from this point.
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