Question

two men on either side of a cliff 80 m high observe the angle of elevation of the top of the cliff to be 30 degree and 60 degree respectively .find the distance between the two men

Answer 1

let h be the height of the cliff and A and B be the point of observation and DC be the cliff
then in ∆ADC
tan 60°= p/b
√3 = 80/b1
b1 = 80/√3. ....(1)

and in ∆BDC
tan 30° = p/b
1/√3 = 80/b2
b2 = 80√3. ......(2)

on adding eq. (1) and(2) we get
b1+b2 = 80/√3+80√3
AB = (80+240)/√3
= 320/√3
= 185 (approx.)

Answer 2

In ∆ABC

tan 30 = AC / BC

1/√3 = 80/y

y = 80√3


In ∆ADC

tan60 = AC/CD

√3 = 80/x

x = 80/√3

$\frac{80}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{80 \sqrt{3} }{3}$

Distance between two men

AD = BC + CD

AD = y + x

AD =

$= 80 \sqrt{3} + \frac{80 \sqrt{3} }{3}$
$= \frac{240 \sqrt{3 } + 80 \sqrt{3} }{3}$

$= \frac{320 \sqrt{3} }{3}$

= 106.6√3 m