Question 32f48

Jul 28, 2016

$x = {72.2}^{\circ}$

$y = {26.6}^{\circ}$

Explanation:

Given

$\cos x + \cos y = 1.2 \ldots . . \left(1\right)$

$\sin x + \sin y = 1.4 \ldots \ldots \left(2\right)$

Squaring and addig (1) and (2)

${\cos}^{2} x + {\sin}^{2} x + {\cos}^{2} y + {\sin}^{2} y + 2 \cos x \cos y + 2 \sin x \sin y = {1.4}^{2} + {1.2}^{2}$

$\implies 2 + 2 \cos \left(x - y\right) = 3.4$

$\implies 2 \cos \left(x - y\right) = 1.4$

$\implies x - y = {\cos}^{-} 1 \left(0.7\right)$

$\implies x - y = {45.6}^{\circ} \ldots \ldots \left(3\right)$

Deviding (2) by (1)

$\frac{\sin x + \sin y}{\cos x + \cos y} = \frac{1.4}{1.2}$

=>(2sin((x+y)/2)cos((x-y)/2))/ (2cos((x+y)/2)cos((x-y)/2))=7/6#

$\implies \tan \left(\frac{x + y}{2}\right) = \frac{7}{6}$

$\implies \frac{x + y}{2} = {\tan}^{-} 1 \left(\frac{7}{6}\right)$

$\frac{x + y}{2} = {49.4}^{\circ}$

$\implies x + y = {98.8}^{\circ}$.......(4)

Adding (3) & (4)

$\implies 2 x = {98.8}^{\circ} + {45.6}^{\circ} = {144.4}^{\circ}$

$\therefore x = {72.2}^{\circ}$

Subtracting (3) from (4)

$2 y = {98.8}^{\circ} - {45.6}^{\circ} = {53.2}^{\circ}$

$\therefore y = {26.6}^{\circ}$