Question
In Figure. find the values of $a, b$ and $c.$
✓
Answer
In figure, In$\triangle\text{PQS},$
$\angle\text{QPS}+\angle\text{PQS}+\angle\text{PSQ}=180^{\circ} [$sum of all the angles of triangle is $180^\circ ]$
$\Rightarrow 55^\circ + 60^\circ + a = 180^\circ $
$\Rightarrow 115^\circ + a = 180^\circ $
$\Rightarrow a = 180^\circ - 65^\circ = 115^\circ $
Now, $a + b = 180^\circ [$linear pair has sum of $180^\circ ]$
$\Rightarrow 65^\circ + b = 180^\circ $
$\Rightarrow b = 180^\circ - 65^\circ = 115^\circ $
In $\triangle\text{PSR},$
$\angle\text{PSR}+\angle\text{SPR}+\angle\text{PRS}=180^{\circ} [$sum of all angles of a triangle is $180^\circ ]$
$\Rightarrow 115^\circ + c + 40^\circ = 118^\circ $
$\Rightarrow 155^\circ + c = 180^\circ $
$\Rightarrow c = 180^\circ - 155^\circ = 25^\circ $
$\angle\text{QPS}+\angle\text{PQS}+\angle\text{PSQ}=180^{\circ} [$sum of all the angles of triangle is $180^\circ ]$
$\Rightarrow 55^\circ + 60^\circ + a = 180^\circ $
$\Rightarrow 115^\circ + a = 180^\circ $
$\Rightarrow a = 180^\circ - 65^\circ = 115^\circ $
Now, $a + b = 180^\circ [$linear pair has sum of $180^\circ ]$
$\Rightarrow 65^\circ + b = 180^\circ $
$\Rightarrow b = 180^\circ - 65^\circ = 115^\circ $
In $\triangle\text{PSR},$
$\angle\text{PSR}+\angle\text{SPR}+\angle\text{PRS}=180^{\circ} [$sum of all angles of a triangle is $180^\circ ]$
$\Rightarrow 115^\circ + c + 40^\circ = 118^\circ $
$\Rightarrow 155^\circ + c = 180^\circ $
$\Rightarrow c = 180^\circ - 155^\circ = 25^\circ $
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