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Question: Given b = 2, c = \(\sqrt { 3 }\) , ∠A = 30<sup>0</sup>, then the in-radius of ∆ABC is...

Given b = 2, c = 3\sqrt { 3 } , ∠A = 300, then the in-radius of ∆ABC is

A

3−12\frac { \sqrt { 3 } - 1 } { 2 }

B

3+12\frac { \sqrt { 3 } + 1 } { 2 }

C

3−14\frac { \sqrt { 3 } - 1 } { 4 }

D

None of these

Answer

3−12\frac { \sqrt { 3 } - 1 } { 2 }

Explanation

Solution

b = 2, c = 3\sqrt { 3 } , ∠A = 300

a = b2+c2−2bccos⁡A\sqrt { b ^ { 2 } + c ^ { 2 } - 2 b c \cos A }= 4+3−2.23⋅32\sqrt { 4 + 3 - 2.2 \sqrt { 3 } \cdot \frac { \sqrt { 3 } } { 2 } }= 1\sqrt { 1 }= 1

⇒ s = 3+32\frac { 3 + \sqrt { 3 } } { 2 }.Also, ∆ =12\frac { 1 } { 2 }bc sin A32\frac { \sqrt { 3 } } { 2 }

⇒ r =13+1=3−12\frac { 1 } { \sqrt { 3 } + 1 } = \frac { \sqrt { 3 } - 1 } { 2 }