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

                

Answers (1)


Given that,
Line XY and MN intersect at O and \anglePOY = 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 \angleMOY = \anglec [vertically opposite angles]
          \anglea + \anglePOY = c
           126^0 =c

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