Some Applications of Trigonometry | Exercise 9.1

Question 8

  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 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

Setup: A statue stands on top of a pedestal. From a single ground point D, the angle of elevation to the top of the pedestal is 45° and to the top of the statue is 60°. Both angles form right-angled triangles with the same base AD.

Angle of Elevation: The angle between the horizontal and the line of sight looking up to a point.

Tangent Ratio: tanθ=OppositeAdjacent\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}} — used to relate vertical heights to the horizontal ground distance.

We will use trigonometric ratios to find the height of the pedestal.

Step 1 — Set up equations

Let the height of the pedestal be AB=hAB = h meters. Let the statue on top of the pedestal be BC=1.6 mBC = 1.6\text{ m} tall. Let DD be the point on the ground from which the angles of elevation are measured, and let the distance AD=xAD = x meters.

The total height of the pedestal and the statue is: AC=AB+BC=h+1.6AC = AB + BC = h + 1.6

In the right-angled triangle ABD\triangle ABD: tan45=ABAD\tan 45^\circ = \frac{AB}{AD} 1=hx1 = \frac{h}{x} x=h— (1)x = h \quad \text{--- (1)}

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

Diagram 1

Step 2 — Solve for the height of the pedestal

Substitute the value of xx from equation (1) into equation (2): 3=h+1.6h\sqrt{3} = \frac{h + 1.6}{h}

Multiply both sides by hh: 3h=h+1.6\sqrt{3}h = h + 1.6

Rearrange the terms to group hh on one side: 3hh=1.6\sqrt{3}h - h = 1.6

Factor out hh: h(31)=1.6h(\sqrt{3} - 1) = 1.6

Solve for hh: h=1.631h = \frac{1.6}{\sqrt{3} - 1}

Rationalize the denominator by multiplying the numerator and denominator by (3+1)(\sqrt{3} + 1): h=1.6(3+1)(31)(3+1)h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}

Apply the algebraic identity (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2 to the denominator: h=1.6(3+1)(3)212h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} h=1.6(3+1)31h = \frac{1.6(\sqrt{3} + 1)}{3 - 1} h=1.6(3+1)2h = \frac{1.6(\sqrt{3} + 1)}{2}

Simplify the expression: h=0.8(3+1)h = 0.8(\sqrt{3} + 1)

If we substitute the value of 31.732\sqrt{3} \approx 1.732: h=0.8(1.732+1)h = 0.8(1.732 + 1) h=0.8(2.732)h = 0.8(2.732)

h2.19 mh \approx 2.19\text{ m}

Answer

The height of the pedestal is 0.8(3+1) m0.8(\sqrt{3} + 1)\text{ m} (or approximately 2.19 m2.19\text{ m}).

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 3m, 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 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 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 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
  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 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 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 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 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 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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