The diameter of the circumcircle is -sin 50-0-76-A- 10-5 cmB- 13-15 cmC- 15-25 cmD- 14-75 cm

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Ratio of Sides of Special Triangles

The diameter ...

Question

In triangle $A B C, ∠ A = 130^{\circ}, B C = 10 c m$ . The diameter of the circumcircle is $[s i n 50^{\circ} = 0.76]$

10.5 cm

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13.15 cm

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15.25 cm

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14.75 cm

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Solution

The correct option is B
13.15 cm

Draw diameter BD and join CD

ABCD in a cyclic quadrilateral so $∠ C D B = 180^{\circ} - 130^{\circ} = 50^{\circ}$

$∠ B C D = 90^{\circ}$ (Angle subtended by diameter on circumference)

In right-angle triangle BCD

$s i n 50^{\circ} = \frac{B C}{B D} \Rightarrow= \frac{B C}{s i n 50^{\circ}} = \frac{10}{0.76} = 13.15 c m$

Hence, diameter $B D = 13.15 c m$

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In triangle ABC,∠A=130∘,BC=10cm. The diameter of the circumcircle is [sin 50∘=0.76]

The correct option is B 13.15 cm Draw diameter BD and join CD ABCD in a cyclic quadrilateral so ∠CDB=180∘−130∘=50∘ ∠BCD=90∘ (Angle subtended by diameter on circumference) In right-angle triangle BCD sin 50∘=BCBD⇒=BCsin 50∘=100.76=13.15cm Hence, diameter BD=13.15cm

In triangle $A B C, ∠ A = 130^{\circ}, B C = 10 c m$ . The diameter of the circumcircle is $[s i n 50^{\circ} = 0.76]$