In right angled triangle ABC - AB -20 m and - CAB -60- Find BC-20-3B- 10 -3C- 10-3D- 20 -3

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

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Byju's Answer

Standard IX

Mathematics

Basics of Geometry

In right angl...

Question

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

\frac{20}{\sqrt{3}}

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10 \sqrt{3}

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\frac{10}{\sqrt{3}}

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20 \sqrt{3}

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Solution

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

In the right angle triangle ABC, AB=20m
$t a n 60^{\circ} = \frac{o p p o s i t e s i d e}{a d j a c e n t s i d e} = \frac{C B}{A B} = \frac{x}{20 m}$
$x = t a n 60^{\circ} \times 20 m$
$t a n 60^{\circ} = \sqrt{3}$
$x = 20 \sqrt{3}$

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

The correct option is D 20√3 In the right angle triangle ABC, AB=20m tan60∘=opposite sideadjacent side=CBAB=x20m x=tan60∘×20m tan60∘=√3 x=20√3

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

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

In the right angle triangle ABC, AB=20m
$t a n 60^{\circ} = \frac{o p p o s i t e s i d e}{a d j a c e n t s i d e} = \frac{C B}{A B} = \frac{x}{20 m}$
$x = t a n 60^{\circ} \times 20 m$
$t a n 60^{\circ} = \sqrt{3}$
$x = 20 \sqrt{3}$