Find the radius of the circle escribed to the triangle ABC (Shown in the figure below) on the side BC if $∠ N A B = 30^{\circ}; ∠ B A C = 30^{\circ}; A B = A C = 5 .$

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$\frac{(10 \sqrt{2} + 5 \sqrt{3} - 5) (2 - \sqrt{3})}{2 \sqrt{2}}$

$\frac{(10 \sqrt{2} + 5 \sqrt{3} + 5)}{2 \sqrt{2}} (2 - \sqrt{3})$

$\frac{(10 \sqrt{2} + 5 \sqrt{3} - 5)}{2 \sqrt{2}} (2 + \sqrt{3})$

$\frac{(10 \sqrt{2} + 5 \sqrt{2} + 1)}{2 \sqrt{3}} (\sqrt{3} - 1)$

answer is B.

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Detailed Solution

Since we need to compute the radius of an escribed circle, we would be needing the length of
all the sides of the given triangle ABC .
From the question, we already know,4B = AC =5.
For finding the length of side BC,let usdraw a line AD which is 'the bisector of angle BAC,as
shown in the figure below

$\begin{array}{l} ∠ B A D = ∠ D A C = 15^{\circ} \\ t h e r e f o r e \sin 15^{\circ} = \frac{B D}{A B} = \frac{B D}{5} and \sin 15^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ t h e r e f o r e B D = 5 \sin 15^{\circ} = \frac{5 (\sqrt{3} - 1)}{2 \sqrt{2}} \\ w e a l s o k n o w t a h t B C = 2 B D \\ B C = \frac{5 (\sqrt{3} - 1)}{\sqrt{3}} \end{array}$
NoW we know that the required radius
$\begin{array}{l} r_{1} = s \tan (\frac{A}{2}) = (\frac{A B + B C + C A}{2}) \tan (\frac{A}{2}) \\ = (\frac{5 + \frac{5 (\sqrt{3} - 1)}{\sqrt{2}} + 5}{2}) (\tan 15^{\circ}) = (\frac{10 \sqrt{2} + 5 \sqrt{3} - 5}{2 \sqrt{2}}) (2 - \sqrt{3}) \end{array}$