# 2. In Fig. 6.14, lines XY and MN intersect at O. If $\angle POY = 90^o$ and a : b = 2 : 3, find c.

M manish painkra

Given that,
Line XY and MN intersect at O and $\angle$POY = $90^0$ also $a:b = 2:3 \Rightarrow b = \frac{3a}{2}$..............(i)

Since XY is a straight line
Therefore, $\\a+b+\angle POY = 180^0\\ a+b = 180^0-90^0 = 90^0$...........(ii)
Thus, from eq (i) and eq (ii), we get
$\\\Rightarrow \frac{3a}{2}+a = 90^0\\$
$\\\Rightarrow a = 36^0\\$
So,  $b = 54^0\\$
Since $\angle$MOY = $\angle$c [vertically opposite angles]
$\angle$a + $\angle$POY = c
$126^0 =c$

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