Question

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

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

Answer 1

Answer: (A)

$\frac { 1 } { 2 }$sec

Answer 2

using sine rule in ∆ABD

⇒ AD = ……….(i)

Also using sine rule in ∆ADC we get

AD = ……….(ii)

from (i) & (ii)

$\frac { \sin B } { \sin C } = \frac { \sin \frac { A } { 3 } } { \sin \frac { 2 A } { 3 } }$ [Q BD = CD]

= $\frac { 1 } { 2 }$sec $\frac { \mathrm { A } } { 3 }$