The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing

Q: Why is the base of the second triangle $h - \frac{\sqrt{3}}{2}$? Shouldn't it be just $h$?

Great question! Remember, you moved horizontally while climbing. You didn't just go up; you moved closer to the mountain. The total horizontal distance from the start was $h$. Since you moved $\frac{\sqrt{3}}{2}$ km closer, the new base is the original distance minus the distance you covered.

Q: Can we use the Sine Rule instead of Tan?

Absolutely! The Sine Rule is a powerful tool. You could define the triangle formed by the summit, the starting point, and the point P. However, using the tangent ratio is often more direct for right-angled triangles. Both methods will lead you to the same beautiful result.

Q: Why do we assume the mountain is a vertical line? Is that realistic?

In the world of JEE physics and mathematics, we often model complex real-world scenarios with idealized geometric shapes. This 'vertical line' assumption allows us to isolate the core trigonometric principles without getting bogged down by the mountain's actual shape. It's the beauty of abstraction!

Question

The angle of elevation of the summit of a mountain from a point on the ground is $45^\circ$. After climbing up one km towards the summit at an inclination of $30^\circ$ from the ground, the angle of elevation of the summit is found to be $60^\circ$. Then the height (in km ) of the summit from the ground is :

Options:

(A) $\frac{1}{\sqrt{3}+1}$

(B) $\frac{1}{\sqrt{3}-1}$

(C) $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

(D) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the flat, dusty ground at point $A$, looking up at the majestic summit of a mountain, which we will call point $S$. We want to find the total vertical height of this mountain, which we will call $h$.

### Phase 1: The First Vantage Point

From our starting point $A$, we look up at the summit. The angle of elevation is given as $45^{\circ}$.

Because $	an 45^{\circ} = 1$, we know that in the right-angled triangle formed by the summit and the ground, the perpendicular height $h$ and the base distance $AB$ must be exactly equal. Thus, $AB = h$.

### Phase 2: The Journey Upward

We walk exactly $1$ km up a slope that has an inclination of $30^{\circ}$ from the ground, finally reaching a new resting point, $P$. To determine our exact position at point $P$, we resolve this $1$ km path into horizontal and vertical components.

The horizontal distance covered is:
$$1 \cdot \cos 30^{\circ} = \frac{\sqrt{3}}{2} 	ext{ km}$$

The vertical height gained is:
$$1 \cdot \sin 30^{\circ} = \frac{1}{2} 	ext{ km}$$

### Phase 3: The Second Vantage Point

From this new vantage point $P$, the new angle of elevation to the summit is $60^{\circ}$. We consider the smaller right-angled triangle formed at the top.

The remaining base of this triangle is the total horizontal distance minus the distance already covered:
$$	ext{Base} = h - \frac{\sqrt{3}}{2}$$

The remaining height of this triangle is the total mountain height minus the height already gained:
$$	ext{Height} = h - \frac{1}{2}$$

### Phase 4: The Algebraic Resolution

We apply the tangent ratio to this upper triangle, where $	an 60^{\circ} = \frac{	ext{Perpendicular}}{	ext{Base}}$. Substituting our values, we obtain:

$$\sqrt{3} = \frac{h - \frac{1}{2}}{h - \frac{\sqrt{3}}{2}}$$

Cross-multiplying to simplify, we get:

$$\sqrt{3} \left( h - \frac{\sqrt{3}}{2} ight) = h - \frac{1}{2}$$

Expanding the left side yields:

$$\sqrt{3}h - \frac{3}{2} = h - \frac{1}{2}$$

Rearranging the terms to isolate $h$:

$$\sqrt{3}h - h = \frac{3}{2} - \frac{1}{2}$$

$$h(\sqrt{3} - 1) = 1$$

Finally, solving for $h$, we find the total height of the mountain:

**$$h = \frac{1}{\sqrt{3} - 1} 	ext{ km}$$**

Select Answer:

Visualized Solution

Initial Setup

First Angle of Elevation

Base Distance Calculation

Climbing the Incline

Horizontal Position of $P$

Vertical Position of $P$

Second Angle of Elevation

Base of the Upper Triangle

Height of the Upper Triangle

Applying Trigonometry at $P$

Cross-Multiplication

Expanding the Equation

Isolating $h$

Final Height of the Summit

The Sigma Insight: Heights and Distances

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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