Question

How can I solve this question

Answer 1

Answer:

The value of x can be 5 or 2

Step-by-step explanation:

In given triangle ABC

Interior Angle A = $x^{2} +3x+20$

Interior Angle B = $x^{2} +14x-5$

Exterior Angle I = $3x^{2} +10x+25$

We know that

Exterior Angle of triangle is summation of corresponding opposite angles

Angle I = Angle A + Angle B

$3x^{2} +10x+25= (x^{2} +3x+20)+(x^{2} +14x-5)$

$3x^{2} +10x+25= 2x^{2} +17x+15$

$1x^{2}-7x+10=0$

$x^{2}-2x-5x+10=0$

$x(x-2)-5(x-2)=0$

$(x-5)(x-2)=0$