Heights & Distances — Analytical & Application Based Questions

An inclined plane AC is prepared with its base AB which is √3 times its vertical height BC. The length of the inclined plane is 15 m. Find:

(a) value of θ.

(b) length of its base AB, in nearest metre.

Answer

(a) According to question,

⇒ AB = $\sqrt{3} \times BC$

⇒ AB = $\sqrt{3}BC$

⇒ BC = $\dfrac{AB}{\sqrt{3}}$

From figure,

⇒ tan θ = $\dfrac{BC}{AB}$

⇒ tan θ = $\dfrac{BC}{\sqrt{3}BC}$

⇒ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, θ = 30°.

(b) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = BC² + AB²

⇒ 15² = $\Big(\dfrac{AB}{\sqrt{3}}\Big)^2$ + AB²

⇒ 225 = $\dfrac{AB^2}{3}$ + AB²

⇒ 225 = $\dfrac{AB^2 + 3AB^2}{3}$

⇒ 225 = $\dfrac{4AB^2}{3}$

⇒ AB² = $\dfrac{225 \times 3}{4}$

⇒ AB² = $\dfrac{675}{4}$

⇒ AB² = 168.75

⇒ AB = $\sqrt{168.75}$

⇒ AB = 12.99 m

Rounding off,

AB = 13 m.

Hence, AB = 13 m.

A cylindrical drum is unloaded from a truck by rolling it down along a wooden plank. The length of the plank is 10 m and it is making an angle of 10° with the horizontal ground. Find the height from which the cylindrical drum was rolled down. Give your answer correct to 3 significant figures.

Answer

Let AC be the plank and BC be the height from where the drum is rolled down.

⇒ sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

⇒ sin 10° = $\dfrac{BC}{AC}$

⇒ 0.1736 = $\dfrac{BC}{10}$

⇒ BC = 1.736 ~ 1.74 m

Hence, height from which the cylindrical drum was rolled down = 1.74 m.

A tree (TS) of height 30 m stands in front of a tall building (AB). Two friends Rohit and Neha are standing at R and N respectively, along the same straight line joining the tree and the building (as shown in the diagram). Rohit, standing at a distance of 150 m from the foot of the building, observes the angle of elevation of the top of the building as 30°. Neha from her position observes that the top of the building and the tree has the same elevation of 60°.

Find the:

(a) height of the building

(b) distance between

1. Neha and the foot of the building
2. Rohit and Neha
3. Neha and the tree
4. building and the tree.

Answer

(a) From figure,

tan 30° = $\dfrac{AB}{AR}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{150} \\[1em] \Rightarrow AB = \dfrac{150}{\sqrt{3}} = \dfrac{150}{1.732} \\[1em] = 86.6 \text{ m}.$

Hence, height of the building = 86.6 m.

(b)

1. From figure,

tan 60° = $\dfrac{AB}{AN}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{\dfrac{150}{\sqrt{3}}}{AN} \\[1em] \Rightarrow AN = \dfrac{150}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AN = \dfrac{150}{3} = 50\text{ m}.$

Hence, distance between Neha and foot of the building = 50 m.

2. From figure,

RN = AR - AN = 150 - 50 = 100 m.

Hence, distance between Rohit and Neha = 100 m.

3. From figure,

tan 60° = $\dfrac{ST}{NT}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{30}{NT} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30\sqrt{3}}{3} \\[1em] \Rightarrow NT = 10\sqrt{3} \\[1em] \Rightarrow NT = 10 \times 1.732 = 17.32 \text{ m}.$

Hence, distance between Neha and the tree = 17.32 m.

(iv) From figure,

AT = AN - NT = 50 - 17.32 = 32.68 m

Hence, distance between building and tree = 32.68 m.