In a triangle-sine of two angles are given .....find cos of 3rd angle

Hello, gurparwaan!

Quote:

$\text{The sines of two angles of triangle are }\tfrac{5}{13}\text{ and }\tfrac{99}{101}$
$\text{Find the cosine of third angle.}$

. . $(1)\;\frac{245}{1313} \quad (2)\;\frac{255}{1313} \quad (3)\;\frac{735}{1313} \quad (4)\;\frac{765}{1313}$

$\text{Let: }\sin A = \frac{5}{13}\;\hdots\text{ then: }\cos A = \frac{12}{13}$

$\text{Let: }\sin B = \frac{99}{101}\;\hdots\text{ then: }\cos B = \frac{20}{101}}$

$\text{We have: }\:A + B + C \:=\:180^o \quad\Rightarrow\quad C \:=\:180^o - (A+B)$

$\text{Then: }\:\cos C \;=\;\cos[180^o - (A+B)] \;=\;-\cos(A+B)$

$\text{Hence: }\:\cos C \;=\;-\cos A\cos B + \sin A\sin B$

. . . = . . . . . . $=\;-\left(\frac{12}{13}\right)$ $\left(\frac{20}{101}\right) + \left(\frac{5}{12}\right)$ $\left(\frac{99}{101}\right)$

. . . . . . . . . . $=\;-\frac{240}{1313} + \frac{495}{1313}$

. . . . . . . . . . $=\;\;\;\frac{255}{1313} \;\;\hdots\;\text{ answer (2)}$