From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60°. The

Q: Wait, why is the angle $\angle PCT = 30^\circ$?

Great observation! In the large right-angled triangle $\triangle CPT$, the sum of all angles must be $180^\circ$. We know $\angle CTP = 90^\circ$ (since the tower is vertical) and $\angle CPT = 60^\circ$ (given as the angle of elevation). Therefore, $\angle PCT = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. It is all about the internal geometry of the triangle!

Q: Why did we draw a horizontal line from the top of the pole to the tower?

That is a classic geometric construction trick! By drawing that horizontal line, we create a rectangle at the base. This allows us to 'lift' the pole's height up to the tower, effectively splitting the tower's total height $h$ into two segments: the bottom part, which is equal to the pole's height ($20$ m), and the top part, which is $h - 20$. This makes the trigonometry much simpler to handle.

Q: Can we use the sine rule instead of tangent?

Absolutely! The sine rule is a powerful tool in any triangle. You could apply it to $\triangle APC$ or $\triangle CPT$. However, for JEE problems, we often prefer the tangent ratio because it directly relates the vertical height to the horizontal base, which is usually the most efficient path to the solution. Stick with the method that minimizes algebraic complexity!

Question

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $60^\circ$. The pole subtends an angle $30^\circ$ at the top of the tower. Then the height of the tower is:

Options:

(A) $15\sqrt{3}$

(B) $20\sqrt{3}$

(C) $20 + 10\sqrt{3}$

(D) 30

Sigma Percentile · Accepted Answer

### The Geometry of Heights: A Trigonometric Journey

Welcome, fellow explorer of the physical world! Today, we are not just solving a math problem; we are constructing a bridge between the abstract world of angles and the tangible reality of towers and poles.

Imagine you are standing on a flat, sun-drenched plain. In front of you stands a pole, a silent sentinel of $20$ meters. Further away, a majestic tower rises, its height $h$ unknown, waiting for us to unveil it. This is the essence of trigonometry—the art of measuring the unmeasurable.

### Phase 1: Visualizing the Setup

Let us ground ourselves. We have a tower $CT$ of height $h$ and a pole $AP$ of height $20$ m. Both are perpendicular to the horizontal ground $PT$.

The problem gives us a key piece of information: the angle of elevation of the tower's top $C$ from the pole's base $P$ is $60^\circ$. This immediately defines a large right-angled triangle, $	riangle CPT$.

In this triangle, the tangent of the angle of elevation relates the height of the tower to the distance between the pole and the tower:

$$	an 60^\circ = \frac{CT}{PT} = \frac{h}{PT}$$

Since we know $	an 60^\circ = \sqrt{3}$, we can rearrange this to find the base distance: $PT = \frac{h}{\sqrt{3}}$. This is our first anchor point in the calculation.

### Phase 2: The Hidden Geometry

Now, here is where the problem tests your spatial intuition. The pole $AP$ subtends an angle of $30^\circ$ at the top of the tower $C$.

This means if you were perched at the very top of the tower, looking down at the top and base of the pole, the angle between those two lines of sight would be $30^\circ$. So, $\angle ACP = 30^\circ$.

But we also know from our earlier analysis of $	riangle CPT$ that $\angle PCT = 30^\circ$. This is a beautiful coincidence! The total angle at the top of the tower, $\angle ACB$, is the sum of these two angles: $\angle ACB = \angle ACP + \angle PCT = 30^\circ + 30^\circ = 60^\circ$.

### Phase 3: The Rectangle Construction

To make sense of the upper part of the tower, let us draw a horizontal line from the top of the pole $A$ to the tower, meeting it at point $B$. This creates a rectangle $APTB$.

Because opposite sides of a rectangle are equal, the length $AB$ is equal to the base distance $PT$, which we already found to be $\frac{h}{\sqrt{3}}$. The vertical segment $BT$ is equal to the pole's height, $20$ m.

Thus, the remaining height of the tower, $CB$, is simply $h - 20$.

### Phase 4: The Final Calculation

We are now left with a smaller right-angled triangle at the top, $	riangle ABC$. We know the angle $\angle ACB = 60^\circ$, the opposite side $AB = \frac{h}{\sqrt{3}}$, and the adjacent side $CB = h - 20$.

Applying the tangent ratio once more:

$$	an 60^\circ = \frac{AB}{CB} = \frac{h/\sqrt{3}}{h - 20}$$

Substituting $\sqrt{3}$ for $	an 60^\circ$, we get:

$$\sqrt{3} = \frac{h/\sqrt{3}}{h - 20}$$

Multiplying both sides by $\sqrt{3}(h - 20)$, we obtain:

$$3(h - 20) = h$$

Expanding this, we get $3h - 60 = h$, which simplifies to $2h = 60$, or $h = 30$ m. The tower stands at exactly **$30$ meters**. It is a perfect, elegant result, born from the simple harmony of triangles.

Animated Solution for Mathematics - Trigonometry: 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:

Select Answer:

Visualized Solution

Visualizing the Setup

Angle of Elevation

Base Distance $P T$

Angle Subtended at Tower Top

Angle $∠ P C T$

Constructing Rectangle $A P T B$

Dimensions of $△ A B C$

Trigonometry in $△ A B C$

Solving for $h$

Final Calculation

The Sigma Insight: Heights and Distances

The Geometry of Heights

Phase 1

Phase 2

Phase 3

Phase 4

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 :

$A B C$ 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 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: