Explanation
Given, measure of one complementary angle is $$=\dfrac{1}{4}\times 90^o=\dfrac{45}{2}^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-$$ $$\dfrac{45}{2}^o$$ $$=\dfrac{180-45}{2}^o$$ $$=\dfrac{135}{2}^o$$ $$=67.5^o.$$
$$\therefore$$ Measure of a complementary angle of $$\dfrac{1}{4}$$ of a right angle $$=67.5^o$$.
We know, two angles are complementary when they add up to $$90^o.$$
Given, measure of one complementary angle is $$(150-a+b)^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-(150-a+b)^o$$
$$=(90-150+a-b)^o=(-60+a-b)^o=(a-b-60)^o.$$
$$\therefore$$ Measure of the complementary angle of $$(150-a+b)^o$$ $$=(a-b-60)^o.$$
Hence, the statement is true and option $$A$$ is correct.
Let's try and prove the exterior angle property for a triangle.
For the given triangle $$XZY$$,
$$\angle 1+\angle 2+\angle XZY=180^{o}$$
Also, $$\angle 3+\angle XZY=180^{o}$$ ......... (Linear pair of angles)
$$\angle 1+\angle 2+\angle XZY=\angle 3 +\angle XZY$$
$$\Rightarrow \angle 3=\angle 1+\angle 2$$ ...... (which is the Exterior angle property).
Therefore, we can say that an exterior angle of a triangle is equal to the sum of the two interior opposite angles. That is, the statement is true.
Hence, option $$A$$ is correct.
We know, two angles whose sum is equal to $$90^o$$ are known as complementary angles.
Also, acute angle is the angle which is greater than $$\displaystyle { 0 }^{ o }$$ and less than $$\displaystyle { 90 }^{ o }$$.
Hence, iftwo angles are complement of each other, then each of them should necessarily be acute angle.
Therefore, option $$C$$ is correct.
Given, measure of one complementary angle is $$ 77^{\circ}$$.
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-77^o$$ $$=13^o.$$
$$\therefore$$ Measure of a complementary angle of $$ 77^{o}=13^o$$.
Therefore, option $$D$$ is correct.
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