Some Applications of Trigonometry | Exercise 9.1

Question 2

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.

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Solution
Understand the Question
  • Let the tree before breaking be represented vertically, with BCBC as the unbroken standing part and ACAC as the broken top part that bends to touch the ground at point AA.
  • The total height of the tree before breaking is given by Total Height=BC+AC\text{Total Height} = BC + AC.
  • In the right-angled triangle ΔABC\Delta \text{ABC} (with ∠B=90∘\angle \text{B} = 90^\circ):
    • Angle made with the ground: ∠BAC=30∘\angle \text{BAC} = 30^\circ
    • Distance from foot to top on ground: AB=8 mAB = 8\text{ m}
  • Use tan⁡30∘\tan 30^\circ to calculate the standing part BCBC, and cos⁡30∘\cos 30^\circ to calculate the broken part ACAC, then sum both lengths.

Step 1 · Find the Standing Part of the Tree

Let ΔABC\Delta \text{ABC} be the right-angled triangle formed, where ∠B=90∘\angle \text{B} = 90^\circ, ∠A=30∘\angle \text{A} = 30^\circ, and AB=8 mAB = 8\text{ m}.Diagram 1

In right ΔABC\Delta \text{ABC}: tan⁡(angle)=Opposite sideAdjacent side\tan(\text{angle}) = \dfrac{\text{Opposite side}}{\text{Adjacent side}}

tan⁡30∘=BCAB13=BC8BC=83 m\begin{aligned} \tan 30^\circ &= \dfrac{BC}{AB} \\[0.6em] \dfrac{1}{\sqrt{3}} &= \dfrac{BC}{8} \\[0.6em] BC &= \dfrac{8}{\sqrt{3}}\text{ m} \end{aligned}

Step 2 · Find the Broken Part of the Tree

In right ΔABC\Delta \text{ABC}: cos⁡(angle)=Adjacent sideHypotenuse\cos(\text{angle}) = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}}

cos⁡30∘=ABAC32=8ACAC=8×23AC=163 m\begin{aligned} \cos 30^\circ &= \dfrac{AB}{AC} \\[0.6em] \dfrac{\sqrt{3}}{2} &= \dfrac{8}{AC} \\[0.6em] AC &= \dfrac{8 \times 2}{\sqrt{3}} \\[0.6em] AC &= \dfrac{16}{\sqrt{3}}\text{ m} \end{aligned}

Step 3 · Calculate the Total Height of the Tree

The total height of the tree is the sum of the standing part BCBC and the broken part ACAC:

Total height=BC+AC=83+163=8+163=243\begin{aligned} \text{Total height} &= BC + AC \\[0.6em] &= \dfrac{8}{\sqrt{3}} + \dfrac{16}{\sqrt{3}} \\[0.6em] &= \dfrac{8 + 16}{\sqrt{3}} \\[0.6em] &= \dfrac{24}{\sqrt{3}} \end{aligned}

Rationalizing the denominator:

Total height=243×33=2433=83 m\begin{aligned} \text{Total height} &= \dfrac{24}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[0.6em] &= \dfrac{24\sqrt{3}}{3} \\[0.6em] &= 8\sqrt{3}\text{ m} \end{aligned}
Answer

83 m8\sqrt{3}\text{ m}

Common Mistakes
  • Incomplete Height: Only finding the vertical standing part BC=83 mBC = \dfrac{8}{\sqrt{3}}\text{ m} and forgetting that the broken bent part ACAC must be added to get the full original height.
  • Incorrect Trig Ratio Value: Confusing tan⁡30∘=13\tan 30^\circ = \dfrac{1}{\sqrt{3}} with tan⁡60∘=3\tan 60^\circ = \sqrt{3}, or taking cos⁡30∘\cos 30^\circ as 12\dfrac{1}{2} instead of 32\dfrac{\sqrt{3}}{2}.
  • Skipping Rationalization: Leaving the final answer as 243 m\dfrac{24}{\sqrt{3}}\text{ m} instead of simplifying it to 83 m8\sqrt{3}\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 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 3 m, 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 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 60∘60^\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 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

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 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 m7\text{ 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 m75\text{ 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 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 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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