Some Applications of Trigonometry | Exercise 9.1

Question 8

A statue, 1.6 m1.6\text{ 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 6060^\circ and from the same point the angle of elevation of the top of the pedestal is 4545^\circ. Find the height of the pedestal.

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Solution
Understand the Question
  • A statue of height 1.6 m1.6\text{ m} stands on top of a vertical pedestal of height hh.
  • From a single observation point on the ground, the angle of elevation to the top of the pedestal is 4545^\circ, and to the top of the statue is 6060^\circ.
  • Both right-angled triangles share the same horizontal base distance xx.
  • By applying the tangent ratio (tanθ=OppositeAdjacent\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}) to both triangles, we get a system of equations to solve for hh.

Step 1 · Set Up Equations Using Trigonometric Ratios

Let the height of the pedestal be AB=h mAB = h\text{ m} and the height of the statue be BC=1.6 mBC = 1.6\text{ m}. Let DD be the observation point on the ground with distance AD=x mAD = x\text{ m}. Total height AC=AB+BC=h+1.6 mAC = AB + BC = h + 1.6\text{ m}.Diagram 1

In right ABD\triangle \text{ABD}: tan45=ABAD\tan 45^\circ = \dfrac{AB}{AD} 1=hx    x=h(1)1 = \dfrac{h}{x} \implies x = h \quad \dots (1)

In right ACD\triangle \text{ACD}: tan60=ACAD\tan 60^\circ = \dfrac{AC}{AD} 3=h+1.6x(2)\sqrt{3} = \dfrac{h + 1.6}{x} \quad \dots (2)

Step 2 · Solve for the Height of the Pedestal

Substitute x=hx = h from (1)(1) into (2)(2): 3=h+1.6h\sqrt{3} = \dfrac{h + 1.6}{h} 3h=h+1.6\sqrt{3}h = h + 1.6 3hh=1.6\sqrt{3}h - h = 1.6 h(31)=1.6h(\sqrt{3} - 1) = 1.6 h=1.631h = \dfrac{1.6}{\sqrt{3} - 1}

Rationalise the denominator:

h=1.6(3+1)(31)(3+1)=1.6(3+1)(3)212=1.6(3+1)31=1.6(3+1)2=0.8(3+1) m\begin{aligned} h &= \dfrac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[0.6em] &= \dfrac{1.6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} \\[0.6em] &= \dfrac{1.6(\sqrt{3} + 1)}{3 - 1} \\[0.6em] &= \dfrac{1.6(\sqrt{3} + 1)}{2} \\[0.6em] &= 0.8(\sqrt{3} + 1)\text{ m} \end{aligned}

Substituting 31.732\sqrt{3} \approx 1.732:

h=0.8(1.732+1)=0.8(2.732)2.19 m\begin{aligned} h &= 0.8(1.732 + 1) \\[0.6em] &= 0.8(2.732) \\[0.6em] &\approx 2.19\text{ m} \end{aligned}
Answer

0.8(3+1) m0.8(\sqrt{3} + 1)\text{ m} (or 2.19 m\approx 2.19\text{ m})

Common Mistakes
  • Opposite Side Error: Using 1.6 m1.6\text{ m} instead of (h+1.6) m(h + 1.6)\text{ m} as the opposite side for tan60\tan 60^\circ. The statue alone does not form a right-angled triangle with the ground base.
  • Angle Association: Swapping the angles—the larger angle (6060^\circ) corresponds to the higher point (top of the statue), while 4545^\circ corresponds to the top of the pedestal.
  • Algebraic Simplification: Forgetting to rationalise 1.631\dfrac{1.6}{\sqrt{3}-1} by multiplying numerator and denominator by the conjugate (3+1)(\sqrt{3}+1).

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 3030^\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 3030^\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 3030^\circ to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 6060^\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 3030^\circ. Find the height of the tower.

Q5

A kite is flying at a height of 60 m60\text{ 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 6060^\circ. Find the length of the string, assuming that there is no slack in the string.

Q6

A 1.5 m1.5\text{ m} tall boy is standing at some distance from a 30 m30\text{ m} tall building. The angle of elevation from his eyes to the top of the building increases from 3030^\circ to 6060^\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 4545^\circ and 6060^\circ respectively. Find the height of the tower.

Q8

A statue, 1.6 m1.6\text{ 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 6060^\circ and from the same point the angle of elevation of the top of the pedestal is 4545^\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 3030^\circ and the angle of elevation of the top of the tower from the foot of the building is 6060^\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 6060^\circ and 3030^\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 6060^\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 3030^\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 m7\text{ m} high building, the angle of elevation of the top of a cable tower is 6060^\circ and the angle of depression of its foot is 4545^\circ. Determine the height of the tower.
Q13
  1. As observed from the top of a 75 m75\text{ m} high lighthouse from the sea-level, the angles of depression of two ships are 3030^\circ and 4545^\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 m1.2\text{ m} tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m88.2\text{ m} from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 6060^\circ. After some time, the angle of elevation reduces to 3030^\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 3030^\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 6060^\circ. Find the time taken by the car to reach the foot of the tower from this point.
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