Question

​let a, b and c be the sides opposite to the angles A, B and C respectively of a triangle ABC. find the value of k such that a + b = kc

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ \mathrm{Using}\mathrm{sine}\mathrm{rule} \\ \frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}=\mathrm{m} \\ \mathrm{So}\mathrm{putting}\mathrm{above}\mathrm{in}\mathrm{a}+\mathrm{b}=\mathrm{kc} \\ \mathrm{k}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{msinA}+\mathrm{msinB}}{\mathrm{msinC}}=\frac{\mathrm{sinA}+\mathrm{sinB}}{\mathrm{sinC}}=\frac{2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos {\displaystyle \frac{\left(A-B\right)}{2}}}{2\sin {\displaystyle \frac{C}{2}}\cos {\displaystyle \frac{C}{2}}} \\ =\frac{\sin \left({\displaystyle 90-\frac{C}{2}}\right)\cos {\displaystyle \frac{\left(A-B\right)}{2}}}{\sin {\displaystyle \frac{C}{2}}\cos {\displaystyle \frac{C}{2}}}=\frac{\cos \left({\displaystyle \frac{C}{2}}\right)\cos {\displaystyle \frac{\left(A-B\right)}{2}}}{\sin {\displaystyle \frac{C}{2}}\cos {\displaystyle \frac{C}{2}}} \\ =\frac{\cos {\displaystyle \frac{\left(A-B\right)}{2}}}{\sin {\displaystyle \frac{C}{2}}} \\ \mathrm{Regards} \end{gathered}$$