Question

Two interior angles of a triangle are in the ratio 3:4. The opposite exterior angle is 140 degrees. Find the measure of all the angles of the triangle.

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

Let us consider a triangle ABC such that side BC is produced to D to form an exterior angle ∠ADC.

According to statement,

• The ratio of interior opposite angles is 3 : 4.

• Let ∠ABC = 3x

and

• Let ∠BAC = 4x

and

• ∠ADC = 140°.

Now,

We know that,

• The exterior angle of a triangle is always equal to the sum of the interior opposite angles.

$\rm :\implies\: \angle \:ADC \: = \: \angle \:BAC \: + \: \angle \:ABC$

$\rm :\implies\:3x + 4x = 140$

$\rm :\implies\:7x = 140$

$\rm :\implies\:x = 20$

So,

$\begin{gathered}\begin{gathered}\bf \: angles \: are \: \:  - \begin{cases} &\sf{\angle \:ABC = 3x = 3 \times 20 = 60} \\ &\sf{\angle \:BAC = 4x = 4 \times 20 = 80} \end{cases}\end{gathered}\end{gathered}$

Now

In triangle ABC,

• We know that, angle sum of a triangle is 180°.

$\rm :\implies\:\angle \:ACB + \angle \:BAC + \angle \:ABC = 180$

$\rm :\implies\:\angle \:ACB + 80 + 60 = 180$

$\rm :\implies\:\angle \:ACB = 180 - 140$

$\rm :\implies\:\angle \:ACB = 40$

Hence,

$\begin{gathered}\begin{gathered}\bf \: angles \:of \: \triangle \: are \: \:  - \begin{cases} &\sf{\angle \:ABC = 60} \\ &\sf{\angle \:BAC = 80} \\ \ &\sf{\angle \:ACB = 40} \end{cases}\end{gathered}\end{gathered}$