At the foot of a mountain, the elevation of the summit is . After ascending 2 kilometers towards the

Q: What is the correct answer for CUET UG Mathematics?

The correct answer is B. Let H be the height of the mountain and X be the horizontal distance from the foot of the mountain to the summit. Initially, the angle of elevation is $45^\circ$. So, $\tan 45^\circ = H/X \Rightarrow 1 = H/X \Rightarrow H = X$. A person ascends 2 km towards the mountain at an incline of $30^\circ$. Horizontal distance covered = $2 \cos 30^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$ km. Vertic

Q: At the foot of a mountain, the elevation of the summit is . After ascending 2 kilometers towards the mountain, at an incline of , the elevation changes to . Determine the height of the mountain?

The correct answer is B. Let H be the height of the mountain and X be the horizontal distance from the foot of the mountain to the summit. Initially, the angle of elevation is $45^\circ$. So, $\tan 45^\circ = H/X \Rightarrow 1 = H/X \Rightarrow H = X$. A person ascends 2 km towards the mountain at an incline of $30^\circ$. Horizontal distance covered = $2 \cos 30^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$ km. Vertic

Q: What are the options for this CUET UG Mathematics question?

A. $(\sqrt{3} - 1)$ kilometre · B. $(\sqrt{3} + 1)$ kilometre · C. $(\sqrt{3} - 2)$ kilometres · D. $(\sqrt{3} + 2)$ kilometres

Question

At the foot of a mountain, the elevation of the summit is . After ascending 2 kilometers towards the mountain, at an incline of , the elevation changes to . Determine the height of the mountain?

FirstInTest · Accepted Answer

Let H be the height of the mountain and X be the horizontal distance from the foot of the mountain to the summit.
Initially, the angle of elevation is $45^\circ$.
So, $	an 45^\circ = H/X \Rightarrow 1 = H/X \Rightarrow H = X$.

A person ascends 2 km towards the mountain at an incline of $30^\circ$.
Horizontal distance covered = $2 \cos 30^\circ = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$ km.
Vertical height gained = $2 \sin 30^\circ = 2 	imes \frac{1}{2} = 1$ km.

After ascending, the new horizontal distance from the base of the mountain to the point directly below the summit is $X - \sqrt{3} = H - \sqrt{3}$.
The new height from the ground is $H - 1$ (since the person is now 1km higher).

At this new position, the angle of elevation to the summit is $60^\circ$.
So, $	an 60^\circ = \frac{	ext{remaining height}}{	ext{remaining horizontal distance}}$
$\sqrt{3} = \frac{H - 1}{H - \sqrt{3}}$
$\sqrt{3}(H - \sqrt{3}) = H - 1$
$\sqrt{3}H - 3 = H - 1$
$\sqrt{3}H - H = 3 - 1$
$H(\sqrt{3} - 1) = 2$
$H = \frac{2}{\sqrt{3} - 1}$

Rationalize the denominator:
$H = \frac{2}{\sqrt{3} - 1} 	imes \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$
$H = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$
$H = \frac{2(\sqrt{3} + 1)}{3 - 1}$
$H = \frac{2(\sqrt{3} + 1)}{2}$
$H = \sqrt{3} + 1$ km.

Answer

$(\sqrt{3} - 1)$ kilometre

Answer

$(\sqrt{3} + 1)$ kilometre

Answer

$(\sqrt{3} - 2)$ kilometres

Answer

$(\sqrt{3} + 2)$ kilometres

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