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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Solution

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

In $Δ A B C, ∠ A = 40^{\circ}, B C = 8 c m .$ Find the circumdiameter of the triangle. $[s i n 40^{\circ} = 0.64]$

In the given triangle $A B C, ∠ A = 50^{\circ}, ∠ c = 50^{\circ}$ , AB =10 cm, then find the length of BC. $[s i n 60^{\circ} = 0.86, s i n 50^{\circ} = 0.76]$

In the given figure triangle ABC is circumscribed by a circle, AB =5 cm, $∠ A = 30^{\circ}, ∠ C = 60^{\circ}$ . Find the length of side BC. $[s i n 60^{\circ} = 0.86, s i n 30^{\circ} = 0.5]$

Q. In

△ A B C

A B = A C = x; B C = 10

cm and the area of the triangle is

60 c m^{2}

. Find

x

(in cm).

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

Draw diameter BD passing through centre O, join CD In Δ BCD, ∠ D=60∘ (Angle subtended by the same chord BC) ∠ BDC=90∘ (Angle subtended by diameter BD on circumference) In right - angled triangle BCD sin 60∘=BCBD=10BD BD=10sin 60=100.87=11.49 cm Hence circumdiameter = 11.49 cm

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