# In a triangle length of two larger sides are 10 cm and 9 cm respectively . If the angles of the triangle are in A.P then the length of the third side can be?

Sep 21, 2017

$7.449 \text{cm, or, "2.551"cm.}$

#### Explanation:

We will follow the Usual Notation for a $\Delta .$

Given that, the angles of a triangle are in A.P.

So, we suppose that, in

DeltaABC, A=x-y, B=x, &, C=x+y.

But,

$A + B + C = {180}^{\circ} \therefore \left(x - y\right) + x + \left(x + y\right) = {180}^{\circ} .$

$\therefore x = {60}^{\circ} .$

Since, $C > B > A , c > b > a .$

$\therefore c = 10 , b = 9.$

Applying the Sine-Rule for $\Delta A B C ,$ we have,

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C .$

$\therefore \frac{a}{\sin} \left(60 - y\right) = \frac{9}{\sin} {60}^{\circ} = \frac{10}{\sin} \left(60 + y\right) \ldots \left[\because , B = x = 60\right] .$

$\therefore \sin \left(60 + y\right) = \frac{10}{9} \cdot \sin 60 \approx 0.96225 .$

$\therefore 60 + y = {74.21}^{\circ} , \mathmr{and} , {180}^{\circ} - {74.21}^{\circ} = {105.79}^{\circ} .$

$\therefore y = {14.21}^{\circ} , \mathmr{and} , {45.79}^{\circ} .$

$\therefore A = 60 - y = {45.79}^{\circ} , \mathmr{and} , {14.21}^{\circ} .$

$\therefore \frac{a}{\sin} \left({45.79}^{\circ}\right) = \frac{9}{\sin} {60}^{\circ} , \mathmr{and} , \frac{a}{\sin} \left({14.21}^{\circ}\right) = \frac{9}{\sin} {60}^{\circ} .$

$\therefore a = \frac{9 \sin {45.79}^{\circ}}{\sin} {60}^{\circ} , \mathmr{and} , a = \frac{9 \sin {14.21}^{\circ}}{\sin} {60}^{\circ} .$

$\therefore a \approx 7.449 \text{cm, or, "a~~2.551"cm.}$