Question

in the figure two straight line PQ and RS intersect each other at O. If POT = 75°. find the value of a, b and c.

Answer 1

take RS line ,
4b+75+b = 180
5b = 180-75=105
b= 105/5 = 21

now take PQ line,
75+b+a =180
75+21+a =180
a = 180-75-21 =84

now again take RS line for uper angle
2c+a = 180
2c = 180-84=96
c = 48

hence,
a=84
b=21
c=48

Answer 2

Answer:

Since OR and OS are in the same line

$\therefore$ $\angle$ROP + $\angle$POT + $\angle$TOS = 180°

4b° + 75° + b° = 180°

5b° + 75° = 180°

5b° = 105° $\implies$ b = 21°

_____________________

Since PQ and RS intersect at O. Therefore,

$\angle$QOS = $\angle$POR (vertically opposite angles)

a = 4b

a = 4 $\times$ 21 = 84

______________________

Now, OR and OS are in the sams line. Therefore,

$\angle$ROQ + $\angle$QOS = 180°

$\rightarrow$ 2c + a = 180

$\rightarrow$ 2c + 84 = 180

$\rightarrow$2c = 96 $\implies$ c = 48

Hence, a = 84, b = 21 and c = 48