Question

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 pedestal is 45'. Find the height of the pedestal.

Answer 1

Correct Question :–

A statue 1.6 m tall, stands on the top of the 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.

Answer :–

The height of the pedestal $= \boxed{ \tt \orange{h = 0.8( \sqrt{3} + 1) m}}$

To Find :–

• Height of the pedestal.

Solution :–

Let the height of pedestal BD be h.

• ∠C = 60°
• ∠D = 45°
• CD = 1.6cm

i.e. ∠CAB = 60°, ∠DAB = 45° and CD = 1.6cm

In the right angled triangle ABD, we have,

→ $\tan(45 \degree) = \dfrac{BD}{AB}$

We know that, tan 45° is 1.

→ $1 = \dfrac{h}{AB}$

→ $AB = h.............(1)$

In right angled triangle ABC, we have,

→ $\tan(60 \degree) = \dfrac{BC}{AB}$

We know that, tan 60° is √3

→ $\sqrt{3} = \dfrac{BD + DC}{AB}$

→ $\sqrt{3} = \dfrac{h + 1.6}{AB}$

→ $ab = \dfrac{h + 1.6}{ \sqrt{3} } .............(2)$

Now,

Comparing the both equations (1) & (2), we get,

→ $h = \dfrac{h + 1.6}{ \sqrt{3} }$

→ $\sqrt{3} h = h + 1.6$

→ $\sqrt{3} h - h = 1.6$

→ $h( \sqrt{3} - 1) = 1.6$

→ $h = \dfrac{1.6}{ \sqrt{3} - 1 } \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1}$

→ $h = \dfrac{1.6( \sqrt{3} + 1) }{ {( \sqrt{3}) }^{2} - {(1)}^{2} }$

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

→ $h = \dfrac{ \cancel{1.6}( \sqrt{3} + 1) }{ \cancel{2} }$

→ $h = \dfrac{0.8( \sqrt{3} +1 )}{1}$

→ $\boxed{ \tt \green{h = 0.8( \sqrt{3} + 1)}}$

Hence,

The height of the pedestal $= \boxed{ \tt \red{h = 0.8( \sqrt{3} + 1) m}}$

Answer 2

Step-by-step explanation:

$\mathfrak{ \huge{ \red{ \underline{hope \: it \: helps}}}}$