Question

Prove the sum of interior angles of a triangle is 180 degree

Answer 1

Answer:

To Prove:-

Sum of interior angles of a triangle is 180°.

Steps:-

Let us draw a ∆ABC.

Then, draw a || line. That should be || to base of the triangle (Side BC).

From our construction, we can tell that Angle (A+D+E) is equal to 180°.

And, Angle (D) = Angle B...(1)

Also, Angle (E) = Angle C...(2)

Hence,

From this we can tell that:

Angle (A+B+C) = 180°

Proof:-

(LONG EXPLAINATION:-)

We know, Degree of one angle of a polygon equals to (formula):

$\boxed{\tt\large{\frac {(n-2) \times {180}^{\circ}}{n}}}$

(Where n is the side of the polygon)

Hence,

In case of a triangle, n will be equal to 3 as their are 3 sides in the triangle.

Hence,

$\tt{\frac {(3-2) \times {180}^{\circ}}{3}}$

(Formation of the equation as per the formula)

$\tt{= \frac {1 \times {180}^{\circ}}{3} }$

(We have Subtracted 3 from 2 that yields 1. This change is done in the nominator)

$\tt{= \frac {{180}^{\circ}}{3}}$

(Multiplied 180° with 1 and the answer is 180°)

$\tt{= \frac {\cancel {{180}^{\circ}}}{\cancel {3}}}$

(Cancelled 180° with 3)

$\boxed{\tt\blue{= {60}^{\circ}}}$

(Therefore, each angle equals 60°)

NOW,

We also know that a triangle consists of three angles. Hence,

$\sf{{60}^{\circ} \times 3}$

$\boxed{\large\sf\blue{={180}^{\circ}}}$

(Sum of all of them).

$\boxed{\text{HENCE, PROVED.}}$

______________________...

SHORTCUT:- (Less Explaination)

{Note: You can also use the Fomula :

$\large\sf{(n-2) \times {180}^{\circ}}$

In the above formula we have divided it with sides to take out the measurement of one angle in the polygon but if we apply directly without Dividing it with it's sides, the total measurement should be the output. This above process is just for explaining}