MCQ
If the complement of an angle is $79^\circ ,$ then the angle will be of:
- A$1^\circ $
- ✓$11^\circ $
- C$79^\circ $
- D$101^\circ $
Let the angle be $x^\circ .$ Then, the complement of $x$ will be $(90 - x)^\circ .$
Given, complement of $x^\circ $ is $79^\circ .$
$\therefore(90-\text{x})^\circ=79^\circ$
$\Rightarrow\text{x}^\circ=90^\circ-79^\circ=11^\circ$
Therefore, the required angle is $11^\circ .$
Note Sum of the complementary angles is $90^\circ .$
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