JEE Main 2023 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: From the top $A$ of a vertical wall $A B$ of height $30 m$ , the angles of depression of the top $P$ and bottom $Q$ of a vertical tower $P Q$ are $1 5^{\circ}$ and $6 0^{\circ}$ respectively, $B$ and $Q$ are on the same horizontal level. If $C$ is a point on $A B$ such that $C B = P Q$ , then the area (in $m^{2}$ ) of the quadrilateral $B C P Q$ is equal to

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Progressive Hints

Hint 1

The Spark: Visualize the wall AB and the tower PQ as two vertical pillars on a flat ground. The angles of depression from the top of the wall A to the top and bottom of the tower create two distinct right-angled triangles. The horizontal distance BQ is the bridge between these two triangles.

Hint 2

The Arsenal: Use the tangent function to relate the vertical heights to the horizontal distance BQ. Remember that the angle of depression from A to Q is equal to the angle of elevation from Q to A due to alternate interior angles. Similarly, use the angle of depression to P to find the height of the tower.

Hint 3

The Execution: Start by calculating the horizontal distance BQ using the triangle ABQ: \tan 60^\circ = \frac{AB}{BQ}. Then, define the height of the tower as h and use the triangle ACP to set up the equation: \tan 15^\circ = \frac{AC}{CP}, where AC = 30 - h and CP = BQ.

Visualized Solution

Visualizing the Wall

Let $A B$ be the vertical wall of height $30 m$ .
Let the ground be a horizontal line where point $B$ lies.

Angle of Depression to $Q$

Point $Q$ is the bottom of the tower on the ground.
Angle of depression from $A$ to $Q$ is $6 0^{\circ}$ .
By alternate interior angles, $∠ A QB = 6 0^{\circ}$ .

Setting up $△ A B Q$

In right-angled $△ A B Q$ :
$tan 6 0^{\circ} = \frac{Perpendicular}{Base} = \frac{A B}{B Q}$

Calculating Distance $B Q$

$3 = \frac{30}{B Q}$
$B Q = \frac{30}{3} = 103 m$

Angle of Depression to $P$

Let $P Q = h$ be the height of the tower.
Angle of depression from $A$ to top $P$ is $1 5^{\circ}$ .
Point $C$ is on $A B$ such that $C B = P Q = h$ .
Draw horizontal $C P$ , so $C P = B Q = 103 m$ .

Analyzing $△ A C P$

In right-angled $△ A C P$ :
$∠ A P C = 1 5^{\circ}$ (Alternate interior angle).
Vertical side $A C = A B - C B = 30 - h$ .

Applying $tan 1 5^{\circ}$

Using tangent in $△ A C P$ :
$tan 1 5^{\circ} = \frac{A C}{C P} = \frac{30 - h}{10 3}$

The Value of $tan 1 5^{\circ}$

Recall standard trigonometric value:
$tan 1 5^{\circ} = 2 - 3$

Setting up the Equation for $h$

Substitute $tan 1 5^{\circ}$ :
$2 - 3 = \frac{30 - h}{10 3}$
Cross-multiply: $30 - h = 103 (2 - 3)$

Solving for Tower Height $h$

Expand the right side:
$30 - h = 203 - 10 (3 \times 3) = 203 - 30$
Rearrange to find $h$ :
$h = 30 + 30 - 203 = 60 - 203 m$

Area of Quadrilateral $B C P Q$

Quadrilateral $B C P Q$ has vertical parallel sides $C B$ and $P Q$ .
Since $C B = P Q = h$ and $∠ B = 9 0^{\circ}$ , $B C P Q$ is a rectangle.
$Area = Base \times Height = B Q \times h$

Substituting Values for Area

$Area = 103 \times (60 - 203)$

Final Calculation

Expand the product:
$Area = 6003 - 200 (3 \times 3)$
$Area = 6003 - 600$
$Area = 600 (3 - 1) m^{2}$

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

We are given a vertical wall

A B

of height

30 m

and a tower

P Q

of height

h

. The points

B

and

Q

lie on the same horizontal ground level.

Our objective is to determine the area of the quadrilateral

B C P Q

The Foundation

Consider the right-angled triangle

A B Q

. The observer at

A

views the base of the tower

Q

with an angle of depression of

6 0^{\circ}

By the property of alternate interior angles, the angle of elevation from

Q

A

is also

6 0^{\circ}

. We apply the tangent ratio:

tan 6 0^{\circ} = \frac{A B}{B Q}

Given

tan 6 0^{\circ} = 3

and

A B = 30

, we solve for the horizontal distance

B Q

3 = \frac{30}{B Q} \Rightarrow B Q = \frac{30}{3} = 103 m

The Tower's Height

Next, we consider the top of the tower

P

. The angle of depression from

A

P

1 5^{\circ}

. Let

C

be a point on

A B

such that

C B = P Q = h

This construction forms a rectangle

B C P Q

, where the horizontal distance

C P = B Q = 103 m

. The vertical segment

A C

is given by:

A C = A B - C B = 30 - h

In the right-angled triangle

A C P

, we apply the tangent function:

tan 1 5^{\circ} = \frac{A C}{C P} = \frac{30 - h}{10 3}

Using the identity

tan 1 5^{\circ} = 2 - 3

, we substitute and solve for

h

2 - 3 = \frac{30 - h}{10 3}

30 - h = 103 (2 - 3) = 203 - 30

h = 60 - 203 m

Final Calculation

The quadrilateral

B C P Q

is a rectangle with base

B Q

and height

h

. The area is calculated as follows:

Area = B Q \times h

Area = 103 \times (60 - 203)

Area = 6003 - 200 (3) = 6003 - 600

Factoring the expression, we obtain the final result:

Area

= 600 (3 - 1) m^{2}

Common Traps & FAQs

Wait, why is the quadrilateral BCPQ a rectangle?

Great observation! Since both the wall AB and the tower PQ are vertical, they are parallel to each other. The segment BQ is on the horizontal ground, and we are given that CB = PQ = h. Because the angle at B is 90 degrees, the quadrilateral BCPQ has two parallel vertical sides of equal length and a horizontal base, which confirms it is a rectangle.

Why can't we just use the sine rule to solve this?

You absolutely could use the sine rule, but it would be like using a sledgehammer to crack a nut! Since we are dealing with right-angled triangles formed by vertical heights and horizontal distances, the tangent function is the most direct and efficient tool. It keeps the algebra clean and minimizes the chance of calculation errors.

How do I calculate tan 15 degrees quickly in an exam?

Memorizing the value of \tan 15^\circ = 2 - \sqrt{3} is a huge time-saver for JEE. If you forget it, don't panic! You can derive it using the formula \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}. Plugging in the values \tan 45^\circ = 1 and \tan 30^\circ = \frac{1}{\sqrt{3}} will lead you to the same result.

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Visualized Solution

Visualizing the Wall

Angle of Depression to $Q$

Setting up $△ A B Q$

Calculating Distance $B Q$

Angle of Depression to $P$

Analyzing $△ A C P$

Applying $tan 1 5^{\circ}$

The Value of $tan 1 5^{\circ}$

Setting up the Equation for $h$

Solving for Tower Height $h$

Area of Quadrilateral $B C P Q$

Substituting Values for Area

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Foundation

The Tower's Height

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

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 :

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:

$A B C$ 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.

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Visualized Solution

Visualizing the Wall

Angle of Depression to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABQ

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

Angle of Depression to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan15∘

The Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan15∘

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

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

Area of Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BCPQ

Substituting Values for Area

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Foundation

The Tower's Height

Final Calculation

Similar Questions

Angle of Depression to $Q$

Setting up $△ A B Q$

Calculating Distance $B Q$

Angle of Depression to $P$

Analyzing $△ A C P$

Applying $tan 1 5^{\circ}$

The Value of $tan 1 5^{\circ}$

Setting up the Equation for $h$

Solving for Tower Height $h$

Area of Quadrilateral $B C P Q$