The sum of exterior angles of a triangle is

180^{\circ}

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270^{\circ}

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360^{\circ}

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540^{\circ}

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Solution

The correct option is C $360^{\circ}$
The sum of exterior angle of a triangle is $360$
the interior angle of a triangle add upto $180$ , For any given corner exterior angle is $180$ minus interior angle.
So sum of exterior angle is (180-a)+(180-b)+(180-c)=3*180-(a+b+c)=540-180= $360^{0}$

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If the side of a triangle are produced, then the sum of the exterior angles i.e. a+b+c is equal to

Q. Question 23
The sum of all exterior angles of a triangle is
a)

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(a) 90°

(b) 180°

(c) 270°

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The sides $B C$ , $C A$ and $A B$ of $Δ A B C$ are produced in order to form exterior angles $∠ A C D$ , $∠ A B G$ , and $∠ C A E$ then $∠ A C D + ∠ C A E + ∠ A B G$ is

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The sum of exterior angles of a triangle is

The correct option is C 360∘The sum of exterior angle of a triangle is 360the interior angle of a triangle add upto 180, For any given corner exterior angle is 180 minus interior angle.So sum of exterior angle is (180-a)+(180-b)+(180-c)=3*180-(a+b+c)=540-180=3600

The correct option is C $360^{\circ}$
The sum of exterior angle of a triangle is $360$
the interior angle of a triangle add upto $180$ , For any given corner exterior angle is $180$ minus interior angle.
So sum of exterior angle is (180-a)+(180-b)+(180-c)=3*180-(a+b+c)=540-180= $360^{0}$