JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be $4 5^{\circ}$ from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be $3 0^{\circ}$ , then the distance (in m) of the foot of the tower from the point A is:

Q: Is it possible to solve this without rationalizing the denominator?

You could leave the answer as $\frac{30\sqrt{3}}{\sqrt{3}-1}$, but in competitive exams like JEE, the options are almost always rationalized. Rationalizing is a standard mathematical hygiene practice that makes the final value easier to compare and interpret.

Select Answer:

Progressive Hints

Hint 1

The Spark: Imagine the tower as a vertical line segment. Point A and the foot of the tower form the base of a right-angled triangle. The key is to visualize that the horizontal distance from the tower to the observer remains constant, whether you are at point A or point B.

Hint 2

The Arsenal: You have two right-angled triangles. The first triangle, with vertex at A, has an angle of

4 5^{\circ}

. The second triangle, with vertex at B, has an angle of

3 0^{\circ}

. Use the tangent function,

tan (θ) = \frac{opposite}{adjacent}

, to relate the height of the tower and the horizontal distance.

Hint 3

The Execution: Let the height of the tower be

h

and the distance be

y

. From A, you get

tan (4 5^{\circ}) = \frac{h}{y}

. From B, you get

tan (3 0^{\circ}) = \frac{h - 30}{y}

. Solve this system of equations by substituting

y = h

into the second equation.

Visualized Solution

Visualizing the Tower and Point $A$

Let the height of the vertical tower be $h$ .
Let the distance of the foot of the tower from point $A$ be $y$ .

Angle of Elevation from $A$

Angle of elevation from point $A$ to the top of the tower is $4 5^{\circ}$ .

Relating $h$ and $y$

In the right-angled triangle formed from $A$ :
$tan (4 5^{\circ}) = \frac{h}{y}$
Since $tan (4 5^{\circ}) = 1$ , we get $h = y$ .

Introducing Point $B$

Point $B$ is $30 m$ vertically above point $A$ .

Angle of Elevation from $B$

Angle of elevation from point $B$ to the top of the tower is $3 0^{\circ}$ .

Geometry from Point $B$

Horizontal distance from $B$ to the tower is $y$ .
Vertical height from the horizontal level of $B$ to the top of the tower is $h - 30$ .

Applying Tangent from $B$

In the upper triangle:
$tan (3 0^{\circ}) = \frac{h - 30}{y}$
Since $tan (3 0^{\circ}) = \frac{1}{3}$ , we have:
$\frac{1}{3} = \frac{h - 30}{y}$

Substituting $y = h$

Substitute $y = h$ into the equation:
$\frac{1}{3} = \frac{h - 30}{h}$

Cross-Multiplication

Cross-multiplying gives:
$h = 3 (h - 30)$
Expanding the bracket:
$h = 3 h - 303$

Isolating $h$

Rearranging the terms to group $h$ :
$303 = 3 h - h$
Factoring out $h$ :
$303 = h (3 - 1)$
Solving for $h$ :
$h = \frac{30 3}{3 - 1}$

Rationalizing the Denominator

To rationalize, multiply numerator and denominator by $(3 + 1)$ :
$h = \frac{30 3 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )}$
The denominator becomes: $(3)^{2} - 1^{2} = 3 - 1 = 2$ .

Final Calculation

Simplify the expression:
$h = \frac{30 3 ( 3 + 1 )}{2} = 153 (3 + 1)$
Distributing $3$ :
$h = 15 (3 + 3)$
Since $y = h$ , the distance is $15 (3 + 3) m$ .

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Let the height of the vertical tower be

h

and the distance from point

A

to the foot of the tower be

y

When we stand at point

A

, we observe the top of the tower at an angle of elevation of

4 5^{\circ}

. This forms a right-angled triangle where the opposite side is

h

and the adjacent side is

y

Using the definition of the tangent function:

tan (4 5^{\circ}) = \frac{h}{y}

Since

tan (4 5^{\circ}) = 1

, we immediately find that

h = y

. This confirms that the distance to the tower is exactly equal to its height.

The Second Observation

Now, we ascend

30 m

vertically to a new point,

B

. From this elevated position, the angle of elevation to the top of the tower changes to

3 0^{\circ}

The horizontal distance remains

y

, but the vertical height of this new triangle is

h - 30

, as we are positioned

30 m

above the ground. Applying the tangent function again:

tan (3 0^{\circ}) = \frac{h - 30}{y}

Given that

tan (3 0^{\circ}) = \frac{1}{3}

, our equation becomes:

\frac{1}{3} = \frac{h - 30}{y}

The Algebraic Bridge

We have a system of two equations with two unknowns. By substituting

h

for

y

in our second equation, we reduce the problem to a single variable:

\frac{1}{3} = \frac{h - 30}{h}

Cross-multiplying yields

h = 3 (h - 30)

, which expands to

h = 3 h - 303

. Rearranging the terms to isolate

h

303 = h (3 - 1)

Solving for

h

, we obtain:

h = \frac{30 3}{3 - 1}

Final Calculation

To reach the final answer, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate,

(3 + 1)

h = \frac{30 3 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )}

The denominator simplifies to

(3)^{2} - 1^{2} = 3 - 1 = 2

. Thus:

h = \frac{30 3 ( 3 + 1 )}{2} = 153 (3 + 1)

Distributing the

3

, we arrive at the final result:

$h = 15 (3 + 3) m$

Since

y = h

, the distance to the tower is $15 (3 + 3) m$ .

Common Traps & FAQs

Wait, why is the horizontal distance from B to the tower the same as from A to the tower?

Because the tower is vertical and the ground is horizontal, the horizontal displacement remains constant as you move vertically up. Imagine a plumb line dropped from B; it hits the ground at A. The distance from that plumb line to the tower is the same as the distance from A to the tower.

Why did we use

tan

instead of

sin

cos

Trigonometry is about choosing the right tool for the job. We know the adjacent side (distance

y

) and we are interested in the opposite side (height

h

). The tangent function is the only one that directly relates these two sides without involving the hypotenuse, which we don't know and don't need.

Is it possible to solve this without rationalizing the denominator?

You could leave the answer as

\frac{30 3}{3 - 1}

, but in competitive exams like JEE, the options are almost always rationalized. Rationalizing is a standard mathematical hygiene practice that makes the final value easier to compare and interpret.

Select Answer:

Visualized Solution

Visualizing the Tower and Point $A$

Angle of Elevation from $A$

Relating $h$ and $y$

Introducing Point $B$

Angle of Elevation from $B$

Geometry from Point $B$

Applying Tangent from $B$

Substituting $y = h$

Cross-Multiplication

Isolating $h$

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Second Observation

The Algebraic Bridge

Final Calculation

Similar Questions

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:

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:

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 :

Select Answer:

Visualized Solution

Visualizing the Tower and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

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

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

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

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

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

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

Cross-Multiplication

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

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Second Observation

The Algebraic Bridge

Final Calculation

Similar Questions

Visualizing the Tower and Point $A$

Angle of Elevation from $A$

Relating $h$ and $y$

Introducing Point $B$

Angle of Elevation from $B$

Geometry from Point $B$

Applying Tangent from $B$

Substituting $y = h$

Isolating $h$