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

• Jul 11th 2011, 06:23 PM
gurparwaan
In a triangle-sine of two angles are given .....find cos of 3rd angle
the sines of two angles of triangle are 5/13 and 99/101 ...find cosine of third angle?
1)245/313
2)255/313
3)735/1313
4)765/1313
• Jul 11th 2011, 10:59 PM
Soroban
Re: In a triangle-sine of two angles are given .....find cos of 3rd angle
Hello, gurparwaan!

Quote:

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

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

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

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

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

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

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

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

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

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