MCQ
The complement of $(90-a)^0$ is:
- A$(90+2 a)^{\circ}$
- B$(90-a)^0$
- ✓$a^{\circ}$
- D$-a^{\circ}$
✓
Answer
Correct option: C.
$a^{\circ}$
We know that the sum of complementary angles are $90^{\circ}$
Let the complementary angle of $(90-\mathrm{a})^{\circ}$ be $x$
$x+(90-a)^{\circ}=90^{\circ}$
$x=90^{\circ}-(90-a)^{\circ}$
$x=90^{\circ}-90^{\circ}+a^{\circ}$
$x=a^{\circ}$.
Let the complementary angle of $(90-\mathrm{a})^{\circ}$ be $x$
$x+(90-a)^{\circ}=90^{\circ}$
$x=90^{\circ}-(90-a)^{\circ}$
$x=90^{\circ}-90^{\circ}+a^{\circ}$
$x=a^{\circ}$.
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