A) \[\frac{15}{16}\]
B) \[\frac{8}{17}\]
C) \[\frac{8}{15}\]
D) \[\frac{1}{2}\]
E) \[\frac{11}{15}\]
Correct Answer: C
Solution :
\[\Delta =2bc-({{b}^{2}}+{{c}^{2}}-{{a}^{2}})=2bc(1-\cos A)\] \[=2bc\,2{{\sin }^{2}}A/2\] ...(i) But \[\Delta =\frac{1}{2}bc\,\sin A\] \[=\frac{1}{2}bc\,2\,\sin \frac{A}{2}\cos \frac{A}{2}\] \[\Delta =bc\,\sin \frac{A}{2}\cos \frac{A}{2}\] ?. (ii) From Eqs. (i) and (ii), we get \[\tan \frac{A}{2}=\frac{1}{4}\] \[\therefore \] \[\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}=\frac{\frac{1}{2}}{1-\frac{1}{16}}=\frac{8}{15}\]
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