Question

In an ambiguous case, If the remaining angles of the triangles

formed with a, b and A be $B _ { 2 } , C _ { 2 }$ then

$\frac { \sin C _ { 1 } } { \sin B _ { 1 } } + \frac { \sin C _ { 2 } } { \sin B _ { 2 } } =$

• A. $2 \cos A$
• B. $\cos A$
• C. $2 \sin A$
• D. sinA

Answer 1

Answer: (A)

$2 \cos A$

Answer 2

In and $A C B _ { 2 }$

$\frac { \sin C _ { 1 } } { \sin B _ { 1 } }$ = $\frac { A B _ { 1 } } { A C } = \frac { c _ { 1 } } { b }$ and $\frac { \sin C _ { 2 } } { \sin B _ { 2 } } = \frac { c _ { 2 } } { b }$

$\therefore$ $\frac { \sin C _ { 1 } } { \sin B _ { 1 } } + \frac { \sin C _ { 2 } } { \sin B _ { 2 } } = \frac { c _ { 1 } + c _ { 2 } } { b }$

$\left[ \begin{array} { l } \because \cos A = \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 b c } \text { or } c ^ { 2 } - ( 2 b \cos A ) c + \left( b ^ { 2 } - a ^ { 2 } \right) = 0 \\ \therefore c _ { 1 } + c _ { 2 } = 2 b \cos A \end{array} \right]$

$\frac { \sin C _ { 1 } } { \sin B _ { 1 } } + \frac { \sin C _ { 2 } } { \sin B _ { 2 } }$=$\frac { 2 b \cos A } { b } = 2 \cos A$.