A circle of radius R is circumscribing a triangle ABC- having - A -30- - BC -20 cm- Then the radius - R - is equal to -sin 30-0-5-A- 20 cmB- 10 cmC- 40 cmD- 20 -3 cm

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Sine Rule

A circle of r...

Question

A circle of radius R is circumscribing a triangle ABC, having $∠ A = 30^{\circ}$ , & BC = 20 cm. Then the radius 'R' is equal to $[s i n 30^{\circ} = 0.5]$

40 cm

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20 cm

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10 cm

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$20 \sqrt{3} c m$

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Solution

The correct option is B
20 cm

Draw Diameter BD and Join CD, BD = 2R

In $△ B D C$ $∠ B C D = 90^{o}$ (angle subtended by the diameter at the circumference)

$s i n 30 = \frac{B C}{B D} = \frac{B C}{2 R}$

$\frac{B C}{s i n 30^{\circ}} = 2 R$

$\frac{20}{s i n 30^{\circ}} = 2 R$

$\frac{20}{0.5} = 2 R$

R =20 cm

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A circle of radius R is circumscribing a triangle ABC, having ∠A=30∘, & BC = 20 cm. Then the radius 'R' is equal to [sin 30∘=0.5]

The correct option is B 20 cm Draw Diameter BD and Join CD, BD = 2R In △BDC ∠BCD=90o (angle subtended by the diameter at the circumference) sin 30=BCBD = BC2R BCsin 30∘=2R 20sin 30∘=2R 200.5=2R R =20 cm

A circle of radius R is circumscribing a triangle ABC, having $∠ A = 30^{\circ}$ , & BC = 20 cm. Then the radius 'R' is equal to $[s i n 30^{\circ} = 0.5]$