Question

Prove that the summ of the angles of triangle in 180

Answer 1

$\mathcal{\huge{\underline{\purple{Question:-}}}}$

Prove that the sum of all angles of triangle is 180°.

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Given :-

• Angle 1, 2 and 3 are angles of the triangle.

To Prove :-

• angle (1 + 2 + 3) = 180°

Construction :-

• Draw a line DE parallel to BC.

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$\mathcal{\huge{\underline{\purple{Solution:-}}}}$

angle (1 + 4 + 5) = 180° ( linear pair )

=> 1 + 2 + 3 = 180°

• angle 2 = angle 4 ( alt. int. angles )

• angle 3 = angle 5 ( alt. int. angles )

=> 1 + 2 + 3 = 180°

Hence, sum of all angles of the triangle is 180°.

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Answer 2

$\bigstar{\mathtt{\huge{\underline{\red{Question\:?}}}}}$

✴ Prove that the sum of the angles of triangle in 180 ?

$\mathcal{\huge{\fbox{\green{Solution:-}}}}$

Given :-

• A given triangle ABC.

To Find :-

• The sum of the angles of triangle in 180.

Construction :-

Draw aline PQ || BC in the triangle ABC.

Proof :-

The sum of the angles of triangle in 180 is the angle sum property of a Δ.

A given ∆ABC.

Here, PQ is a straight line. So by the figure, The

sum of angles around a point is 360°.

∠PAB + ∠BAC + ∠QAC = 180° ………(1)

Since, PQ||BC and AB, AC are transversals,

Here, ∠QAC = ∠ACB (a pair of alternate angle)

∠PAB = ∠CBA (a pair of alternate angle)

Putting in equation ( 1 )

∠PAB + ∠BAC + ∠QAC = 180° ………(1)

➡ ∠ACB + ∠BAC + ∠CBA= 180°

Thus, the sum of the interior angles of a triangle is 180°.

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