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in Co-ordinate geometry by (63.5k points)

If in a triangle ABC, a = 1 + 3cm, b = 2cm and  ∠C = 60°, then find the other two angles and the third side. 

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Given that,

a = 1 + √3 , b = 2 and ∠C =  60º

We have, c2 = a2 + b2 – 2ab cos C

⇒ c2 = (1+ √3)2 + 4 – 2(1 + √3) · 2 cos60º

⇒ c2 = 1 + 2√3 + 3 + 4 – 2 – 2√3

⇒ c2 = 6 ⇒ c = √6 

Using sine rule,

a/sinA = b/sinB = c/sinC

⇒ (1 + √3)/sinA = 2/sinB = 6/sin60º

∴sin B = 2sin60º/6 = (2 x √3/2)/√6 = 1/2

∴ ∠B = 45º

⇒ ∠A = 180º – (60º + 45º) = 75º

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