In - ABC - - A -60- BC -10 cm- The circumdiameter of the triangle is -sin 60-0-87-A- 12-11 cmB- 14-97 cmC- 11-49 cmD- 13-45 cm

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Ratio of Sides of Special Triangles

In Δ ABC , ∠ ...

Question

In $Δ A B C, ∠ A = 60^{\circ}, B C = 10 c m .$ The circumdiameter of the triangle is $[s i n 60^{\circ} = 0.87]$

12.11 cm

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11.49 cm

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13.45 cm

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14.97 cm

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Solution

The correct option is B
11.49 cm

Draw diameter BD passing through centre O, join CD

In $Δ B C D, ∠ D = 60^{\circ}$ (Angle subtended by the same chord BC)

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

In right - angled triangle BCD

$s i n 60^{\circ} = \frac{B C}{B D} = \frac{10}{B D}$

$B D = \frac{10}{s i n 60} = \frac{10}{0.87} = 11.49 c m$

Hence circumdiameter = 11.49 cm

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Similar questions

In $Δ A B C, ∠ A = 60^{\circ}, B C = 10 c m .$ Find the circumdiameter of the triangle. $[s i n 60^{\circ} = 0.87]$

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In Δ ABC, ∠ A=60∘, BC=10 cm. The circumdiameter of the triangle is [sin 60∘=0.87]

In $Δ A B C, ∠ A = 60^{\circ}, B C = 10 c m .$ The circumdiameter of the triangle is $[s i n 60^{\circ} = 0.87]$