Question

determine height of a mountain if the angle of its top of an unknown distance from this is 30° and at a distance 10 kilometres for further off from the mountain along the same line the angle of elevation is 15°( take tan15°=0.27)



find it please urgently.....

Answer 1

SOLUTION:

[Look at the attached image]

Let AB = h km

AB = height of the mountain.

Now,

Let C be a point at a distance of x km from the base of the mountain such that ∠ACB = 30°.

Also, let D be any point at a distance of 10 km from C along the same line.

Then,

∠ADB = 15°

And

AD = AC + DC = (x + 10) km

So,

In right-angled ΔBAC,

$tan\ 30^{o}=\frac{P}{B} =\frac{AB}{AC}$

$\implies \frac{1}{\sqrt{3}}=\frac{h}{x}\ \ \ \bf [\therefore tan\ 30^{o} = \frac{1}{\sqrt{3}}]$

$\implies x = h\sqrt{3}$  -------(i)

Now,

In right-angled triangle BAD,

$tan\ 15^{o} = \frac{AB}{AD}$

$\implies 0.27 = \frac{h}{x+10}$

[As given tan 15° = 0.27]

$\implies 0.27(x+10) = h$  ---------(ii)

Now,

By putting the value of x = $\sqrt{3}h$ from the equation(i) and equation(ii), we get,

$\implies 0.27(\sqrt{3}h+10) = h$

$\implies 0.27 \times \sqrt{3}h+ 0.27 \times 10 = h$

$\implies h(1-0.27 \times \sqrt{3})= 0.27 \times 10$

$\implies h(1-0.27 \times 1.732)= 2.7$

$\implies h(1- 0.47) = 2.7$

$\implies 0.53h = 2.7$

$\implies h = \frac{2.7}{0.53} = 5.09 = \boxed{5\ km}$

Hence,

The height of the mountain is 5 km.

Answer 2

$\huge \pink{ \boxed{answer}}$

SOLUTION:

[Look at the attached image]

Let AB = h km

AB = height of the mountain.

Now,

Let C be a point at a distance of x km from the base of the mountain such that ∠ACB = 30°.

Also, let D be any point at a distance of 10 km from C along the same line.

Then,

∠ADB = 15°

And

AD = AC + DC = (x + 10) km

So,

In right-angled ΔBAC,

The height of the mountain is 5 km.