Question

(a) Show that the sum of all angles of a triangle is equal to 180⁰​

Answer 1

Answer:

Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°.

Proof: 

Consider a ∆ABC, as shown in the figure below. To prove the above property of triangles, draw a line PQ←→ parallel to the side BC of the given triangle.

Since PQ is a straight line, it can be concluded that:

∠PAB + ∠BAC + ∠QAC = 180°  ………(1)

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

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

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

Substituting the value of ∠QAC and∠PAB in equation (1),

∠ACB + ∠BAC + ∠CBA= 180°

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

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Answer 2

✯ $\mid$ QUESTION :

• (a) Show that the sum of all angles of a triangle is equal to 180⁰.

✯ $\mid$TO PROOF :

• That $\sf \bold \angle$ ABC + $\sf \bold \angle$CAB + $\sf \bold \angle$ACB = 180°

✯ $\mid$BASED ON THE THEOREM :

• Interior angles of a triangle is 180°.

✯ $\mid$CONSTRUCTION :

• Through point A draw the line i.e., $\sf \: EF \sf \parallel BC$

✯ $\mid$PROOF :

We know that,

• Alternate angles are equal.

HENCE,

$\sf \to \angle ABC = \bold \angle EAB$ [Alternative angles are equal]

$\sf \to \angle BCA = \bold \angle FAC$[Alternative angles are equal]

$\sf \to \bold \angle EAB + \bold \angle BAC + \bold \angle FAC = 180°$ [ sum of all Liners angles = $\sf \bold \angle$ A]

NOW,

By substituting $\sf \bold \angle EAB =\bold \angle ABC$ And, $\sf \bold \angle FAC = \bold \angle BCA$

Finally we can say,

$\sf \bold { \bold {\angle ABC + \angle BCA + \angle CAB = 180° }}$

Or,

• Sum of all interior angles of a triangle is 180°.

$\sf { \pink {\underline{\underline{\sf \purple {HENCE, \: PROVED }}}}}$