The base of a cliff is circular. From the extremities of a diameter of the base angles of elevation of the top of the cliff are 30° and 60° . If the height of the cliff be 500 m, then the diameter of the base of the cliff is

Question

The base of a cliff is circular. From the extremities of a diameter of the base angles of elevation of the top of the cliff are 30° and 60° . If the height of the cliff be 500 m, then the diameter of the base of the cliff is (A)  (2000/√3)m  (B)  (1000/√3)m  (C) (2000/√2) m (D)  1000√3 m

Tardigrade Admin · Accepted Answer

In Δ AEC tan 60°=(500/d1) ⇒ d1=(500/√3) m ... (i) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/08a3dec2d78e721851dd03ed8086fae9-.png" /> and in Δ B E C, tan 30°=(500/d2) ⇒ d2=500 √3 m ... (ii) ∴ Required diameter, A B=d1+d2 =(500/√3)+500 √3=(2000/√3) m

Q. The base of a cliff is circular. From the extremities of a diameter of the base angles of elevation of the top of the cliff are $3 0^{\circ}$ and $6 0^{\circ}$ . If the height of the cliff be $500 m$ , then the diameter of the base of the cliff is

$\frac{2000}{3} m$

$\frac{1000}{3} m$

$\frac{2000}{2} m$

$10003 m$

In $Δ A EC$
$tan 6 0^{\circ} = \frac{500}{d _{1}}$
$\Rightarrow d_{1} = \frac{500}{3} m$ ... (i)

and in $Δ BEC,$
$tan 3 0^{\circ} = \frac{500}{d _{2}}$
$\Rightarrow d_{2} = 5003 m$ ... (ii)
$∴$ Required diameter, $A B = d_{1} + d_{2}$
$= \frac{500}{3} + 5003 = \frac{2000}{3} m$

Q. The base of a cliff is circular. From the extremities of a diameter of the base angles of elevation of the top of the cliff are 30∘ and 60∘ . If the height of the cliff be 500m, then the diameter of the base of the cliff is

3​2000​m

3​1000​m

2​2000​m

10003​m