Question
An angle is greater than $45^{\circ}$. Is its complementary angle greater than $45^{\circ}$ or equal to $45^{\circ}$ or less than $45^{\circ}$ ?
✓
Answer
Let us assume that the two angles are $x$ and $y$
Since, $x$ and $y$ are complementary pair of angles and $x$ is greater $45^{\circ}$
Therefore, $x+y=90^{\circ}$
$y=90^{\circ}-x^{\circ}$
$\text { As } x^{\circ}>45^{\circ}$
$-x^{\circ}<-45^{\circ}$
Add $90^{\circ}$ on both sides to get,
$90^{\circ}-x^{\circ}<90^{\circ}-45^{\circ}$
$y^{\circ}<90^{\circ}-45^{\circ}$
$y^{\circ}<45^{\circ}$
Hence,
Angle $y$ will be less than $45^{\circ}$.
Since, $x$ and $y$ are complementary pair of angles and $x$ is greater $45^{\circ}$
Therefore, $x+y=90^{\circ}$
$y=90^{\circ}-x^{\circ}$
$\text { As } x^{\circ}>45^{\circ}$
$-x^{\circ}<-45^{\circ}$
Add $90^{\circ}$ on both sides to get,
$90^{\circ}-x^{\circ}<90^{\circ}-45^{\circ}$
$y^{\circ}<90^{\circ}-45^{\circ}$
$y^{\circ}<45^{\circ}$
Hence,
Angle $y$ will be less than $45^{\circ}$.
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