Question

A pole 6 m high casts a shadow 2√ 3 m long on the ground, then from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.​

Answer 1

Answer:

In triangle ABC,

tan θ = BC/AB

= 6/2√ 3

= 3/√ 3

= (√ 3 × √ 3 )/ √ 3

= √ 3

tan θ = tan 60°

θ = 60°

Hence, the Sun’s elevation is 60°.

Answer 2

1. Correct Question :

A pole 6m high casts a shadow 2√3m long on the ground, Find the sun's elevation.

AnsWer :

60°.

Solution :

• Height of pole is 6m.
• Shadow casts on ground 2√3m.
• Figure provide above.

So,

$\tt \tan \theta = \frac{perpendicular}{hypotenuse}$

$\tt\implies \tan \theta = \frac{ \cancel6}{ \cancel2 \sqrt{3} }$

$\tt\implies \tan \theta = \frac{3}{ \sqrt{3} }$

Rationalize the denominator,

$\tt\implies\frac{3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \implies \sqrt{3} .$

$\tt\implies \tan \theta = \sqrt{3} .$

$\tt\implies \tan\theta = \tan 60\degree$

$\tt\implies\theta = 60 \degree.$

Therefore, the Sun's elevation be 60°.

$\rule{200}3$

2. Correct Question :

The angle of elevation of top of a tower from two distinct points S and t from foot are complementary. Prove that the height of the tower is √St.

Solution :

• Side BD is (S) on ground.
• Side BC is (t) on ground.
• Height side AB (H).
• Angle ADB is 90- theta.
• Angle ACB is theta.
• Figure provide above.

Now,In ∆ ABD,

$\tt\implies\tan(90 - \theta) = \frac{h}{s} \\ \tt\implies \cot \theta = \frac{h}{s} - - - (1).$

Again, In ∆ ABC,

$\tt \implies\tan \theta = \frac{h}{t} - - - (2).$

Equation 1 Multiply 2.

$\tt\implies \tan \theta \times \cot \theta = \frac{h}{s} \times \frac{h}{t}$

$\tt\implies1 = \frac{ {h}^{2} }{st} .$

$\tt\implies{h}^{2} = st.$

$\tt\implies h = \sqrt{st}$

Hence Proved.