At the foot of a mountain, the elevation of the summit is 45°. After ascending 2 kilometers towards

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 point directly below the summit. Initial observation: Angle of elevation = 45°. $ an 45^{\circ} = H/X \Rightarrow 1 = H/X \Rightarrow H = X$. After ascending 2 km at an incline of 30°: Let the new position be P. The vertical height gained is $2 \sin 30^{\circ} = 2 \times (1/2) = 1$ km. The h

Q: At the foot of a mountain, the elevation of the summit is 45°. After ascending 2 kilometers towards the mountain, at an incline of 30°, the elevation changes to 60°. Determine the height of the…

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 point directly below the summit. Initial observation: Angle of elevation = 45°. $ an 45^{\circ} = H/X \Rightarrow 1 = H/X \Rightarrow H = X$. After ascending 2 km at an incline of 30°: Let the new position be P. The vertical height gained is $2 \sin 30^{\circ} = 2 \times (1/2) = 1$ km. The h

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 45°. After ascending 2 kilometers towards the mountain, at an incline of 30°, the elevation changes to 60°. 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 point directly below the summit.

Initial observation:
Angle of elevation = 45°.
$	an 45^{\circ} = H/X \Rightarrow 1 = H/X \Rightarrow H = X$.

After ascending 2 km at an incline of 30°:
Let the new position be P. The vertical height gained is $2 \sin 30^{\circ} = 2 	imes (1/2) = 1$ km.
The horizontal distance covered is $2 \cos 30^{\circ} = 2 	imes (\sqrt{3}/2) = \sqrt{3}$ km.

From the new position P, the height to the summit is $H-1$.
The horizontal distance from P to the point directly below the summit is $X - \sqrt{3}$.

New angle of elevation = 60°.
$	an 60^{\circ} = \frac{H-1}{X-\sqrt{3}}$
$\sqrt{3} = \frac{H-1}{X-\sqrt{3}}$

Substitute $X=H$ from the first observation:
$\sqrt{3} = \frac{H-1}{H-\sqrt{3}}$
$\sqrt{3}(H-\sqrt{3}) = H-1$
$\sqrt{3}H - 3 = H-1$
$\sqrt{3}H - H = 3-1$
$H(\sqrt{3}-1) = 2$
$H = \frac{2}{\sqrt{3}-1}$

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

The height of the mountain is $(\sqrt{3}+1)$ kilometers.

Answer

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

Answer

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

Answer

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

Answer

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

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