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Q8:
A statue, $1.6$ m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60°$ and from the same point the angle of elevation of the top of the pedestal is $45°$. Find the height of the pedestal.

Solution :

Given:

  • Height of the statue ($AB$) = $1.6$ m.
  • The statue stands on a pedestal ($BC$).
  • Angle of elevation from a point $D$ on the ground to the top of the statue ($A$) is $\angle ADB = 60^\circ$.
  • Angle of elevation from the same point $D$ to the top of the pedestal ($B$) is $\angle CDB = 45^\circ$.

To Find:

The height of the pedestal ($BC = h$ meters).

A B C D Ground 1.6m h

Step 1: Define Variables and Assumptions

Let $BC = h$ be the height of the pedestal in meters.

Let $CD = x$ be the distance from the point $D$ to the base of the pedestal $C$ in meters.

The total height of the statue and pedestal is $AC = AB + BC = 1.6 + h$.

Step 2: Analyze the smaller triangle $\triangle BCD$

In the right-angled triangle $\triangle BCD$, the angle $\angle CDB = 45^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(45^\circ) = \frac{BC}{CD}$

Since $\tan(45^\circ) = 1$:

$1 = \frac{h}{x} \implies x = h$ --- (Equation 1)

Step 3: Analyze the larger triangle $\triangle ACD$

In the right-angled triangle $\triangle ACD$, the angle $\angle ADB = 60^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(60^\circ) = \frac{AC}{CD}$

Substitute $AC = 1.6 + h$ and $CD = x$:

$\sqrt{3} = \frac{1.6 + h}{x}$ --- (Equation 2)

Step 4: Solve for $h$

Substitute $x = h$ (from Equation 1) into Equation 2:

$\sqrt{3} = \frac{1.6 + h}{h}$

Multiply both sides by $h$:

$h\sqrt{3} = 1.6 + h$

Rearrange the terms to isolate $h$:

$h\sqrt{3} - h = 1.6$

$h(\sqrt{3} - 1) = 1.6$

$h = \frac{1.6}{\sqrt{3} - 1}$

Step 5: Rationalize the denominator

To simplify, multiply the numerator and denominator by the conjugate $(\sqrt{3} + 1)$:

$h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$h = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{1.6(\sqrt{3} + 1)}{2}$

$h = 0.8(\sqrt{3} + 1)$

Using $\sqrt{3} \approx 1.732$:

$h = 0.8(1.732 + 1) = 0.8(2.732) = 2.1856$ m.

Final Answer: The height of the pedestal is $0.8(\sqrt{3} + 1)$ m or approximately $2.186$ m.


More Questions from Class 10 Mathematics Applications of Trigonometry EXERCISE 9.1


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