Question

If the median AD of triangle ABC divides the angle BAC in ratio 2 : 1, then $\frac { \sin B } { \sin C }$ is equal to-

• A. $\frac { 1 } { 2 }$sec
• B. $\frac { 1 } { 2 }$ cos
• C. $\frac { 1 } { 2 }$ cosec $\frac { \mathrm { A } } { 3 }$
• D. None

Answer 1

Answer: (D)

None

Answer 2

Using sine rule is ∆ABD

$\frac { \sin 2 \mathrm {~A} / 3 } { \mathrm { BD } }$ = $\frac { \sin B } { A D }$ ............(i)

Using sine rule in ∆ABD

$\frac { \sin C } { A D }$ = ............(ii)

from (i) & (ii)

$\frac { B D \sin B } { \sin 2 \mathrm {~A} / 3 }$ = $\frac { C D \sin C } { \sin A / 3 }$ ⇒ $\frac { \sin B } { \sin C }$ = $\frac { \sin 2 \mathrm {~A} / 3 } { \sin \mathrm { A } / 3 }$

= 2 cos A/3