In right angled triangle ABC - AB -20 m and - CAB -60- - Find BC-A- 20- - 2 mB- 10 -3 mC- 10-3 mD- 20 -3 m

The correct option is D 20√(3) m Let BC=x In the right angle triangle ABC, AB=20 m tan 60^∘=opposite side/adjacent side=CB/AB=x/20 ⇒ x=tan 60^∘× 20 ∵tan 60^∘= ...

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Standard X

Mathematics

Properties of 30°-60°-90° Triangles

In right angl...

Question

In right angled triangle ABC, $A B = 20 m$ and $∠ C A B = 60^{\circ}$ . Find BC.

$\frac{20}{\sqrt{3}} m$

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$10 \sqrt{3} m$

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$\frac{10}{\sqrt{3}} m$

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$20 \sqrt{3} m$

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Solution

The correct option is D
$20 \sqrt{3} m$

Let $B C = x$

In the right angle triangle ABC, $A B = 20 m$
$tan 60^{\circ} = \frac{opposite side}{adjacent side} = \frac{C B}{A B} = \frac{x}{20}$
$\Rightarrow x = tan 60^{\circ} \times 20$
$∵ tan 60^{\circ} = \sqrt{3}$
$∴ x = 20 \sqrt{3} m$

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In right angled triangle ABC, AB=20 m and ∠CAB=60∘. Find BC.

The correct option is D 20√3 m Let BC=x In the right angle triangle ABC, AB=20 m tan60∘=opposite sideadjacent side=CBAB=x20 ⇒x=tan60∘×20 ∵tan60∘=√3 ∴x=20√3 m

In right angled triangle ABC, $A B = 20 m$ and $∠ C A B = 60^{\circ}$ . Find BC.

The correct option is D
$20 \sqrt{3} m$

Let $B C = x$

In the right angle triangle ABC, $A B = 20 m$
$tan 60^{\circ} = \frac{opposite side}{adjacent side} = \frac{C B}{A B} = \frac{x}{20}$
$\Rightarrow x = tan 60^{\circ} \times 20$
$∵ tan 60^{\circ} = \sqrt{3}$
$∴ x = 20 \sqrt{3} m$