Chapter 9: Applications of Trigonometry

An air traffic control tower is 50 m high. From the top of the tower, the angle of depression of a car parked on the ground is 30°.

1. Find the distance of the car from the base of the tower.
   (A) 50 m

   (B) $50\sqrt{3}$ m

   (C) $50/\sqrt{3}$ m

   (D) 100 m
   Show Solution
   Solution: (B) $50\sqrt{3}$ m
   Step 1: Height $h=50$. Angle of elevation from car is also 30°.
   Step 2: $\tan 30^\circ = \frac{50}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{d} \Rightarrow d = 50\sqrt{3}$ m.
2. Another car is observed at an angle of depression of 45°. What is its distance from the base?
   (A) 50 m

   (B) $50\sqrt{2}$ m

   (C) $50\sqrt{3}$ m

   (D) 25 m
   Show Solution
   Solution: (A) 50 m
   Step 1: $\tan 45^\circ = \frac{50}{d'} \Rightarrow 1 = \frac{50}{d'} \Rightarrow d' = 50$ m.
3. If both cars are on the same side of the tower, find the distance between them.
   (A) $50(\sqrt{3}+1)$ m

   (B) $50(\sqrt{3}-1)$ m

   (C) 50 m

   (D) 100 m
   Show Solution
   Solution: (B) $50(\sqrt{3}-1)$ m
   Step 1: Distance $= d - d' = 50\sqrt{3} - 50 = 50(\sqrt{3}-1)$ m.

A boy is flying a kite. The string is 100 m long and makes an angle of 30° with the horizontal ground.

1. Find the height of the kite from the ground (assuming no slack in string).
   (A) 50 m

   (B) $50\sqrt{3}$ m

   (C) 100 m

   (D) $100\sqrt{3}$ m
   Show Solution
   Solution: (A) 50 m
   Step 1: $\sin 30^\circ = \frac{h}{100} \Rightarrow \frac{1}{2} = \frac{h}{100} \Rightarrow h = 50$ m.
2. If the angle changes to 60° keeping the height same, what will be the new length of string required?
   (A) $100\sqrt{3}$ m

   (B) $50\sqrt{3}$ m

   (C) $100/\sqrt{3}$ m

   (D) 50 m
   Show Solution
   Solution: (C) $100/\sqrt{3}$ m
   Step 1: Height $h=50$. $\sin 60^\circ = \frac{50}{L} \Rightarrow \frac{\sqrt{3}}{2} = \frac{50}{L} \Rightarrow L = \frac{100}{\sqrt{3}}$ m.

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

1. Find the height of the remaining part of the tree (standing part).
   (A) $8\sqrt{3}$ m

   (B) $8/\sqrt{3}$ m

   (C) 8 m

   (D) 16 m
   Show Solution
   Solution: (B) $8/\sqrt{3}$ m
   Step 1: $\tan 30^\circ = \frac{y}{8} \Rightarrow \frac{1}{\sqrt{3}} = \frac{y}{8} \Rightarrow y = \frac{8}{\sqrt{3}}$.
2. Find the length of the broken part.
   (A) $16\sqrt{3}$ m

   (B) $16/\sqrt{3}$ m

   (C) 16 m

   (D) 8 m
   Show Solution
   Solution: (B) $16/\sqrt{3}$ m
   Step 1: $\cos 30^\circ = \frac{8}{x} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{x} \Rightarrow x = \frac{16}{\sqrt{3}}$.
3. Find the total height of the tree before it broke.
   (A) $8\sqrt{3}$ m

   (B) $24\sqrt{3}$ m

   (C) 24 m

   (D) $16\sqrt{3}$ m
   Show Solution
   Solution: (A) $8\sqrt{3}$ m
   Step 1: Total $= x+y = \frac{16}{\sqrt{3}} + \frac{8}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ m.