A circle is inscribed in a triangle ABC touches the side AB at D such that AD = 5 and BD = 3. If ∠A = 60° then the length of BC equals

Engineering

Mathematics

Circumradius Inradius and exradius

Question

$\frac{120}{13}$

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Given A = 60°; tan 30° = $\frac{r}{5}$ ⇒ r = $\frac{5}{\sqrt{3}}$

now, $\tan \frac{B}{2} = \frac{r}{3} = \frac{5}{3 \sqrt{3}} (a = ?)$

$cosB = \frac{1 - \tan^{2} (B / 2)}{1 + \tan^{2} (B / 2)} = \frac{1 - (25 / 27)}{1 + (25 / 27)} = \frac{2}{52} = \frac{1}{26}$

Hence sin B = $\frac{15 \sqrt{3}}{26}$

sin C = sin(A + B) = sin A cos B + cos A sin B

$= \frac{\sqrt{3}}{2} \cdot \frac{1}{26} + \frac{1}{2} \cdot \frac{15 \sqrt{3}}{26} = \frac{1}{52} [16 \sqrt{3}] = \frac{4 \sqrt{3}}{13} = sinC$

$\frac{c}{sinC} = \frac{a}{sinA}; a = \frac{8 \cdot \sqrt{3}}{2} \cdot \frac{13}{4 \sqrt{3}}$ = 13