Question

An exterior angle of a triangle is greater than either of the interior opposite angles.​

Answer 1

Answer:

the exterior angle would be greater than either of the opposite angles because an exterior angle is the sum of 2 angles of a triangle thus meaning that the exterior angle is made of the sum of one of the opposite angles, thus it has to be greater than either of the opposite angles

Step-by-step explanation:

Answer 2

Given:

An exterior angle of a triangle is greater than either of the interior opposite angles.

To find:

Whether this statement is true or false?

Solution:

In ∆ABC:

$\rm{\angle ACB +\angle x + \angle y ={180}^{\circ}}$

Now , at point C:

$\rm{\therefore\angle ACD = {180}^{\circ} - \angle ACB}$

$\rm{=>\angle ACD = {180}^{\circ} - \bigg\{{180}^{\circ}-(\angle x +\angle y)\bigg\}}$

$\rm{=>\angle ACD = \cancel{{180}^{\circ}} - \cancel{{180}^{\circ}}+\angle x +\angle y}$

$\rm{=>\angle ACD = \angle x + \angle y}$

$\rm{=>\angle z = \angle x + \angle y}$

Since $\angle x\: and\: \angle y$ are positive , so $\angle z$ will be greater than each of them.

Hence, we can say that:

$\boxed{ \bf{ \angle z > \angle x \: \: and \: \: \angle z > \angle y}}$

Hence, the statement is true.