Why is the base of the second triangle $h-40\sqrt{3}$?

Think of the horizontal distance. Initially, you were $h$ meters away from the foot of the hill. By walking up the slope, you moved $40\sqrt{3}$ meters closer to the hill horizontally. Therefore, your new horizontal distance from the foot of the hill is the original distance $h$ minus the horizontal displacement $40\sqrt{3}$.

JEE Main 2020 - 6 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be $4 5^{\circ}$ . After walking a distance of 80 meters towards the top, up a slope inclined at an angle of $3 0^{\circ}$ to the horizontal plane, the angle of elevation of the top of the hill becomes $7 5^{\circ}$ . Then the height of the hill (in meters) is

Q: Wait, why is the height of the triangle $h-40$ and not just $h$?

Great question! When you move from the ground to point $P$, you are climbing a slope. You are no longer at the base level. Since you've gained 40 meters of altitude, the remaining vertical distance to the top $T$ is the total height $h$ minus your current elevation of 40 meters.

Q: Is there a faster way to calculate $\tan(75^\circ)$?

Absolutely! While you can derive it using $\tan(45^\circ + 30^\circ)$, memorizing $\tan(75^\circ) = 2 + \sqrt{3}$ is a massive time-saver in JEE exams. It's a standard value that appears frequently in geometry problems involving slopes.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. You are not just moving closer to the hill; you are climbing. The observer's position changes in both horizontal and vertical coordinates. Think of the hill's top as a fixed point

T

and the two observation points as vertices of two distinct right-angled triangles sharing the same vertical line.

Hint 2

Hint 2: You need to resolve the 80-meter displacement along the

3 0^{\circ}

slope into its components. Use

x = d cos (θ)

and

y = d sin (θ)

to find how much closer you are to the hill horizontally and how much higher you are vertically.

Hint 3

Hint 3: Once you have the new coordinates, set up the tangent ratio for the second triangle. The equation will look like

tan (7 5^{\circ}) = \frac{h - y _{n e w}}{h - x _{n e w}}

. Use the compound angle formula

tan (4 5^{\circ} + 3 0^{\circ})

to evaluate

tan (7 5^{\circ})

Visualized Solution

Visualizing the Hill and Observation Point

Let the height of the hill be $h$ meters.
Let $F$ be the foot of the hill and $T$ be its top, so $T F = h$ .
Let $A$ be the initial point of observation on the horizontal plane.
The angle of elevation from $A$ to $T$ is $∠ T A F = 4 5^{\circ}$ .

Relating Base and Height

In right-angled $△ T A F$ :
$tan (4 5^{\circ}) = \frac{T F}{A F}$
Since $tan (4 5^{\circ}) = 1$ , we have $1 = \frac{h}{A F}$
Therefore, $A F = h$ .

Moving along the $3 0^{\circ}$ Slope

Distance walked $A P = 80$ meters.
The slope is inclined at an angle of $3 0^{\circ}$ to the horizontal.
Let $P$ be the new position after walking $80$ meters.

Calculating the Coordinates of Point $P$

Vertical height of $P$ from horizontal plane: $y_{P} = 80 sin (3 0^{\circ}) = 80 \times \frac{1}{2} = 40$ meters.
Horizontal distance of $P$ from $A$ : $x_{P} = 80 cos (3 0^{\circ}) = 80 \times \frac{3}{2} = 403$ meters.

Setting up the Elevation from Point $P$

New angle of elevation from $P$ to $T$ is $7 5^{\circ}$ .
Height of $T$ above $P = h - 40$ .
Horizontal distance from $P$ to $T = A F - x_{P} = h - 403$ .

Calculating $tan 7 5^{\circ}$

$tan (7 5^{\circ}) = tan (4 5^{\circ} + 3 0^{\circ}) = \frac{t a n ( 4 5 ^{\circ} ) + t a n ( 3 0 ^{\circ} )}{1 - t a n ( 4 5 ^{\circ} ) t a n ( 3 0 ^{\circ} )}$
$tan (7 5^{\circ}) = \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = \frac{3 + 1}{3 - 1}$
Rationalizing: $\frac{( 3 + 1 ) ^{2}}{3 - 1} = \frac{3 + 1 + 2 3}{2} = 2 + 3$ .

Forming the Equation for $h$

From the observation at $P$ :
$tan (7 5^{\circ}) = \frac{h - 40}{h - 40 3}$
Substituting the value: $2 + 3 = \frac{h - 40}{h - 40 3}$

Solving the Linear Equation

Cross-multiplying: $(2 + 3) (h - 403) = h - 40$
Expanding the left side: $2 h - 803 + 3 h - 40 (3) = h - 40$
Simplifying: $2 h + 3 h - 803 - 120 = h - 40$

Grouping $h$ Terms

Rearranging terms: $(2 h + 3 h - h) = 803 + 120 - 40$
Simplifying: $h (1 + 3) = 803 + 80$
Factoring the right side: $h (1 + 3) = 80 (3 + 1)$

Final Result

Dividing both sides by $(3 + 1)$ :
$h = 80$
Key Takeaway: The height of the hill is $80$ meters.

00:00 / 00:00

The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine standing on a flat, sun-drenched plain, gazing up at a majestic hill. You are at point

A

, and the summit is at point

T

. The angle of elevation is

4 5^{\circ}

In the world of trigonometry, a

4 5^{\circ}

angle is a gift. It tells us that the triangle formed by the hill's height

h

and the horizontal distance

A F

is isosceles.

Since

tan (4 5^{\circ}) = \frac{T F}{A F} = 1

, we immediately know that

A F = h

. The base and the height are identical twins.

The Journey Upward

We walk 80 meters up a slope inclined at

3 0^{\circ}

. We must resolve this 80-meter path into its components.

Using the power of vectors, the vertical gain is:

80 sin (3 0^{\circ}) = 40 meters

The horizontal approach is:

80 cos (3 0^{\circ}) = 403 meters

You have effectively shifted your coordinate system. Your new position,

P

, is 40 meters above the ground and

403

meters closer to the hill's base.

The New Perspective

Now, look up again from point

P

. The angle of elevation is

7 5^{\circ}

. The vertical side of this new triangle is

h - 40

, and the horizontal side is

h - 403

We invoke the tangent ratio:

tan (7 5^{\circ}) = \frac{h - 40}{h - 40 3}

To solve this, we apply the compound angle formula

tan (4 5^{\circ} + 3 0^{\circ}) = \frac{t a n ( 4 5 ^{\circ} ) + t a n ( 3 0 ^{\circ} )}{1 - t a n ( 4 5 ^{\circ} ) t a n ( 3 0 ^{\circ} )}

. This yields the elegant value:

tan (7 5^{\circ}) = 2 + 3

The Algebraic Climax

Now, we equate the expressions:

2 + 3 = \frac{h - 40}{h - 40 3}

Cross-multiplying gives us:

(2 + 3) (h - 403) = h - 40

Expanding this expression:

2 h - 803 + 3 h - 120 = h - 40

Grouping the

h

terms on the left:

h (2 + 3 - 1) = 120 + 803 - 40

This simplifies to:

h (1 + 3) = 80 + 803

Factoring the right side gives

80 (1 + 3)

. The terms cancel perfectly, leaving us with: $h = 80$ meters

The height of the hill is exactly 80 meters. You have conquered the geometry, mastered the trigonometry, and navigated the algebra.

Common Traps & FAQs

Wait, why is the height of the triangle

h - 40

and not just

h

Great question! When you move from the ground to point

P

, you are climbing a slope. You are no longer at the base level. Since you've gained 40 meters of altitude, the remaining vertical distance to the top

T

is the total height

h

minus your current elevation of 40 meters.

Why is the base of the second triangle

h - 403

Think of the horizontal distance. Initially, you were

h

meters away from the foot of the hill. By walking up the slope, you moved

403

meters closer to the hill horizontally. Therefore, your new horizontal distance from the foot of the hill is the original distance

h

minus the horizontal displacement

403

Is there a faster way to calculate

tan (7 5^{\circ})

Absolutely! While you can derive it using

tan (4 5^{\circ} + 3 0^{\circ})

, memorizing

tan (7 5^{\circ}) = 2 + 3

is a massive time-saver in JEE exams. It's a standard value that appears frequently in geometry problems involving slopes.

Enter Numerical Value:

Visualized Solution

Visualizing the Hill and Observation Point

Relating Base and Height

Moving along the $3 0^{\circ}$ Slope

Calculating the Coordinates of Point $P$

Setting up the Elevation from Point $P$

Calculating $tan 7 5^{\circ}$

Forming the Equation for $h$

Solving the Linear Equation

Grouping $h$ Terms

Final Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Journey Upward

The New Perspective

The Algebraic Climax

Similar Questions

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $6 0^{\circ}$ . The pole subtends an angle $3 0^{\circ}$ at the top of the tower. Then the height of the tower is:

If the angle of elevation of a cloud from a point $P$ which is 25 m above a lake be $3 0^{\circ}$ and the angle of depression of reflection of the cloud in the lake from $P$ be $6 0^{\circ}$ , then the height of the cloud (in meters) from the surface of the lake is :

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B (= a)$ subtends an angle of $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is

$A B C$ is a triangular park with $A B = A C = 100$ m. A television tower stands at the midpoint of $B C$ . The angles of elevation of the top of the tower at $A, B, C$ are $4 5^{\circ}, 6 0^{\circ}, 6 0^{\circ}$ , respectively. Find the height of the tower.

$P QR$ is a triangular park with $P Q = P R = 200$ m. A T.V. tower stands at the mid-point of $QR$ . If the angles of elevation of the top of the tower at $P, Q$ and $R$ are respectively $4 5^{\circ}, 3 0^{\circ}$ and $3 0^{\circ}$ , then the height of the tower (in m) is :

Enter Numerical Value:

Visualized Solution

Visualizing the Hill and Observation Point

Relating Base and Height

Moving along the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 30∘ Slope

Calculating the Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Setting up the Elevation from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan75∘

Forming the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Solving the Linear Equation

Grouping .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h Terms

Final Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Journey Upward

The New Perspective

The Algebraic Climax

Similar Questions

Moving along the $3 0^{\circ}$ Slope

Calculating the Coordinates of Point $P$

Setting up the Elevation from Point $P$

Calculating $tan 7 5^{\circ}$

Forming the Equation for $h$

Grouping $h$ Terms