Question

the length of the shadow of a tower is 20m when the altitude of the sun is 60°. what's the height of the tower​

Answer 1

✬ Height = 20√3 or 34.64 m ✬

Step-by-step explanation:

Given:

• Length of shadow of tower is 20 m.
• Altitude of sun is 60°.

To Find:

• What is the height of the tower ?

Solution: Let AB be a tower of h m and BC be its shadow.

In ∆ABC we have

• AB {perpendicular} = h m

• BC {base} = 20 m

• ∠ACB {angle of elevation} = 60°

Applying tanθ

➮ tanθ = Perpendicular/Base

➮ tan60° = AB/BC

➮ √3 = h/20

➮ 20√3 = h

If we take value of √3 to be 1.732 then approx height will be 20 × 1.732 = 34.64 m.

Let's make it by using cotθ

➯ cotθ = Base/Perpendicular

➯ cot60° = BC/AB

➯ 1/√3 = 20/h

➯ h = 20√3

➯ h = 20 × 1.732 = 34.64 m

Hence, the height of tower is 20√3 m or 34.64 m.

Answer 2

$\huge\bold{\mathbb{QUESTION}}$

The length of the shadow of a tower is $\sf 20\:m$ when the altitude of the sun is $\sf 60\degree.$ What's the height of the tower?

$\huge\bold{\mathbb{GIVEN}}$

• The length of the shadow of a tower is $\sf 20\:m.$

• The altitude of the sun is $\sf 60\degree.$

$\huge\bold{\mathbb{TO\:FIND}}$

The height of the tower.

$\huge\bold{\mathbb{DIAGRAM}}$

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$\huge\bold{\mathbb{SOLUTION}}$

Let,

• the height of the tower be $\sf h\:m.$

• $\sf AB$ is the height of the tower.

• $\sf BC$ is the shadow of the tower.

In $\sf \triangle ABC,$

• $\sf AB = h\:m$

• $\sf BC = 20\:m$

• $\sf \angle BAC = 60\degree$

Using $\sf \bold {tan \theta \longrightarrow}$

$\sf tan \theta= {\Large{\frac{AB}{BC}}}$

$\implies\sf tan60\degree= {\Large{\frac{h}{20}}}$

$\implies\sf \sqrt{3}= {\Large{\frac{h}{20}}}$

Multiply both sides with $\sf 20.$

$\implies\sf \sqrt{3}\times20= {\Large{\frac{h}{20}}}\times20$

$\implies\sf 20 \sqrt{3}= {\Large{\frac{h}{\cancel{20}}}}\times{\cancel{20}}$

$\implies\sf 20 \sqrt{3}= h$

$\implies\sf h= 20 \sqrt{3}$

$\implies\sf h=(20\times1.732)\:\:\:\:\{\sqrt{3}=1.732\}$

$\implies \boxed{\sf h = 34.64}$

Alternative method $\longrightarrow$

Using $\sf \bold{cot \theta \longrightarrow}$

$\sf cot \theta= {\Large{\frac{BC}{AB}}}$

$\implies\sf cot60\degree= {\Large{\frac{20}{h}}}$

$\implies\sf {\Large{\frac{1}{\sqrt{3}}}}= {\Large{\frac{h}{20}}}$

By cross multiplication.

$\implies\sf (1\times h)= (20\times{\sqrt{3}})$

$\implies\sf h= 20 \sqrt{3}$

$\implies\sf h= (20\times1.732)\:\:\:\:\{\sqrt{3}=1.732\}$

$\implies \boxed{\sf h = 34.64}$

$\huge\bold{\mathbb{HENCE}}$

$\sf h = 34.64$

Height of the tower $\sf =h\:m = 34.64\:m$

$\huge\bold{\mathbb{THEREFORE}}$

The height of the tower is $\sf 34.64\:m.$

$\huge\bold{\mathbb{DONE}}$