A) \[\frac{15}{16}\]
B) \[\frac{8}{15}\]
C) \[\frac{8}{17}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
\[\Delta =2bc-({{b}^{2}}+{{c}^{2}}-{{a}^{2}})\] \[\Delta =2bc(1-\cos A)=2bc.2{{\sin }^{2}}\frac{A}{2}\] .....(i) \[\Delta =\frac{1}{2}bc\sin A=\frac{1}{2}(bc)2\sin \frac{A}{2}\cos \frac{A}{2}\] \[\Delta =bc\sin \frac{A}{2}\cos \frac{A}{2}\] .....(ii) \ \[\tan \frac{A}{2}=\frac{1}{4}=t\], {by (i) and (ii)} \[\tan A=\frac{2t}{1-{{t}^{2}}}=\frac{8}{15}\].
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