JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $cot^{- 1} (32)$ and $csc^{- 1} (22)$ respectively, then the height of the tower (in metres) is :

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Visualized Solution

Visualizing the Park and Tower

Given: $△ A B C$ is isosceles with $A B = A C = 100$ m.
Let $M$ be the midpoint of $B C$ .
A vertical tower of height $h$ is situated at $M$ . Let the top of the tower be $T$ .
Therefore, $M T = h$ and $M T ⊥$ plane of $△ A B C$ .

Analyzing Elevation from Point $A$

Angle of elevation at $A$ is $α = cot^{- 1} (32)$ .
This implies $cot α = 32$ .
In right $△ T M A$ , the tower is perpendicular to the base.

Expressing Base $A M$ in terms of $h$

In right $△ T M A$ :
$cot α = \frac{A M}{M T} = \frac{A M}{h}$
$\Rightarrow A M = h cot α = 32 h$

Analyzing Elevation from Point $B$

Angle of elevation at $B$ is $β = csc^{- 1} (22)$ .
This implies $csc β = 22$ .
Consider the right $△ T M B$ .

Calculating $cot β$

Using the identity $cot^{2} β = csc^{2} β - 1$ :
$cot^{2} β = (22)^{2} - 1 = 8 - 1 = 7$
$\Rightarrow cot β = 7$

Expressing Base $B M$ in terms of $h$

In right $△ T M B$ :
$cot β = \frac{B M}{M T} = \frac{B M}{h}$
$\Rightarrow B M = h cot β = 7 h$

The Geometry of the Base Triangle

In isosceles $△ A B C$ , $M$ is the midpoint of base $B C$ .
Property: The median to the base of an isosceles triangle is perpendicular to the base.
$∴ A M ⊥ B C \Rightarrow ∠ A M B = 9 0^{\circ}$ .

Applying Pythagoras Theorem

In right $△ A M B$ on the ground plane:
$A M^{2} + B M^{2} = A B^{2}$
This equation links our 3D height $h$ to the 2D base dimensions.

Substituting the Values

Substitute $A M = 32 h$ , $B M = 7 h$ , and $A B = 100$ :
$(32 h)^{2} + (7 h)^{2} = 10 0^{2}$

Simplifying the Equation

$(9 \times 2) h^{2} + 7 h^{2} = 10000$
$18 h^{2} + 7 h^{2} = 10000$
$25 h^{2} = 10000$

Solving for Height $h$

$h^{2} = \frac{10000}{25}$
$h^{2} = 400$
$h = 400 = 20$ m

Conclusion and Key Takeaway

Final Answer: The height of the tower is $20$ metres.
Key Takeaway: In 3D geometry problems, project vertical triangles onto the base plane and use properties like the isosceles median to link the variables.

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine standing in the center of a perfectly symmetrical triangular park,

△ A B C

. The sides

A B

and

A C

are both

100

meters long, forming an isosceles triangle.

A vertical tower of height

h

rises from the midpoint

M

of the base

B C

. The top of the tower is point

T

, and since the tower is vertical,

M T ⊥

ground.

The Vertical Perspective

From point

A

, the angle of elevation to the top

T

α

, where

α = cot^{- 1} (32)

. In the right-angled triangle

△ T M A

cot α = \frac{A M}{M T} = \frac{A M}{h} = 32

This yields the expression for the median:

A M = 32 h

Next, consider point

B

with an angle of elevation

β = csc^{- 1} (22)

. Using the identity

cot^{2} β = csc^{2} β - 1

cot^{2} β = (22)^{2} - 1 = 8 - 1 = 7

Thus,

cot β = 7

. In the right-angled triangle

△ T M B

B M = h cot β = 7 h

The Grounded Reality

We now return to the flat ground of the park. Because

△ A B C

is isosceles with

A B = A C

, the median

A M

to the base

B C

is also the altitude.

Therefore,

∠ A M B = 9 0^{\circ}

. We can now apply the Pythagorean theorem to the right-angled triangle

△ A M B

A M^{2} + B M^{2} = A B^{2}

Final Calculation

Substituting our expressions for

A M

B M

, and the known length

A B = 100

into the Pythagorean equation:

(32 h)^{2} + (7 h)^{2} = 10 0^{2}

Expanding the terms:

(18) h^{2} + 7 h^{2} = 10000

25 h^{2} = 10000

h^{2} = 400

Taking the positive square root, we find the height of the tower: $h = 20$ meters

Common Traps & FAQs

Why is

△ A M B

a right-angled triangle? The problem doesn't explicitly state that

∠ A M B = 9 0^{\circ}

That is a fantastic observation! In geometry, we must rely on the properties of the shapes given. Since

△ A B C

is isosceles with

A B = A C

, the median

A M

drawn from the vertex

A

to the base

B C

is a well-known property that it acts as the altitude as well. Therefore,

A M ⊥ B C

, which confirms that

∠ A M B = 9 0^{\circ}

. This is a classic JEE-style hidden geometric constraint!

Can I use the Law of Cosines instead of the Pythagorean theorem?

You absolutely could, but it would be like taking a long, winding road when there is a superhighway right next to you! The Pythagorean theorem is the most direct and elegant path here because we have already established that

△ A M B

is a right-angled triangle. Using the Law of Cosines would introduce unnecessary complexity and increase the chance of calculation errors.

Why do we assume the tower is at the midpoint

M

We don't assume it; the problem explicitly states it! The tower is situated at the midpoint of

B C

. This is crucial because it defines the geometry of the two vertical triangles

△ T M A

and

△ T M B

. Without this midpoint information, we wouldn't be able to link the height

h

to the base dimensions

A M

and

B M

so cleanly.

Select Answer:

Visualized Solution

Visualizing the Park and Tower

Analyzing Elevation from Point $A$

Expressing Base $A M$ in terms of $h$

Analyzing Elevation from Point $B$

Calculating $cot β$

Expressing Base $B M$ in terms of $h$

The Geometry of the Base Triangle

Applying Pythagoras Theorem

Substituting the Values

Simplifying the Equation

Solving for Height $h$

Conclusion and Key Takeaway

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Vertical Perspective

The Grounded Reality

Final Calculation

Similar Questions

$A B C$ 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 :

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

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:

A pole stands vertically inside a triangular park $A B C$ . Let the angle of elevation of the top of the pole from each corner of the park be $\frac{π}{3}$ . If the radius of the circumcircle of $Δ A B C$ is 2, then the height of the pole is equal to:

If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ , on a line leading to the foot of the tower, are $3 0^{\circ}, 4 5^{\circ}$ and $6 0^{\circ}$ respectively, then the ratio, $A B : B C$ , is:

Select Answer:

Visualized Solution

Visualizing the Park and Tower

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

Expressing Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

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

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

Expressing Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BM in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

The Geometry of the Base Triangle

Applying Pythagoras Theorem

Substituting the Values

Simplifying the Equation

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

Conclusion and Key Takeaway

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Vertical Perspective

The Grounded Reality

Final Calculation

Similar Questions

Analyzing Elevation from Point $A$

Expressing Base $A M$ in terms of $h$

Analyzing Elevation from Point $B$

Calculating $cot β$

Expressing Base $B M$ in terms of $h$

Solving for Height $h$