In the figure, $a$ is greater than $b$ by one third of a right-angle. Find the values of $a$ and $b$."

Given:

$a$ is greater than $b$ by one third of a right-angle.

To do:

We have to find the values of $a$ and $b$.

Solution:

$a = b + \frac{1}{3}(90^o)$

$a = b + 30^o$........(i)

From the figure,

$\angle AOC + \angle BOC = 180^o$              (Linear pair)

$a + b =180^o$…......(ii)

Substituting (i) in (ii), we get,

$b+30^o+b=180^o$

$2b=180^o-30^o$

$b=\frac{150^o}{2}$

$b=75^o$

$\Rightarrow a = 75^o+30^o=105^o$

Hence, $a = 105^o$ and $b = 75^o$.

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Updated on: 10-Oct-2022

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