If $x, y$ and $z$ are exterior angles of $△ A B C$ , then find the value of $x + y + z$ .

x > z > y

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z < x < y

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z > y > x

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z < y < x

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Solution

The correct option is D $z < y < x$
Given, sides $A B > A C > B C$
This means, the angles opposite to the sides will also have the same relation.
$=> ∠ A C B > ∠ A B C > ∠ B A C$ -- (1)
Since, $x^{o}, y^{o}, z^{o}$ are the exterior angles, they will be $180^{o} - ∠ B A C; 180^{o} - ∠ A B C, 180^{o} - ∠ A C B$ respectively. -- (2)
Hence, from $1, 2$
$=> 180^{o} - z^{o} > 180^{o} - y^{o} > 180^{o} - x^{o}$
$=> z^{o} < y^{o} < x^{o}$

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