Question

The angle of elevation of top of a hill is 30°from point on ground on walking 1km towards the hill,angle is founded to be 45°calculate the height of hill

Answer 1

Answer:

1km=1000m

tan45°=h/x

=>h=x----1

tan30°=h/100+x

=>1/√3=h/1000+x

=>√3h=1000+x

=>√3x=1000+x (h=x)

=>√3x-x=1000

=>x(√3-1)=1000

=>x=1000/√3-1

=>x=1000×√3+1/√3-1×√3+1

=>x=1000(√3+1)/3²-1²

=>x=500(√3+1) answer

Answer 2

Refer to the attachment for diagram

In Δ ACB ,

$\tt \implies \tan(45) = \frac{AB}{x}$

$\tt \implies x = AB$

Now , in Δ ADC

$\tt \implies \tan(30) = \frac{AB}{1 + x}$

$\tt \implies \frac{1}{ \sqrt{3} } = \frac{AB}{1 + x}$

$\tt \implies1 + AB = AB \sqrt{3} \: \: \{ \because \: x = AB\}$

$\tt \implies1 = AB( \sqrt{3} - 1)$

$\tt \implies AB = \frac{1}{ \sqrt{3 } - 1}$

$\tt \implies AB = \frac{1}{1.732 - 1}$

$\tt \implies AB = 1.36 \: \: km$

Therefore , the height of hill is 1.36 km