Question

Question 109

In the given figures (i) and (ii), find the values of a, b and c.

Answer 1

In figure (i), $\angle A+\angle B+\angle C={180}^{\circ }$ [sum of all angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {90}^{\circ }+a+{70}^{\circ }={180}^{\circ }$ [$\because \angle A={90}^{\circ }$ and $\angle C={70}^{\circ }$, from the figure]
$\Rightarrow a+{160}^{\circ }={180}^{\circ }\Rightarrow a={180}^{\circ }-{160}^{\circ }={20}^{\circ }$
Since, c is an exterior angle of $\Delta ABD$.
$\therefore \angle c=a+{30}^{\circ }={20}^{\circ }+{30}^{\circ }={50}^{\circ }$
[$\because$ exterior angle = the sum of opposite interior angles]
Also, b is an exterior angle of $\Delta ADC$
$\therefore \angle b={60}^{\circ }+{70}^{\circ }={130}^{\circ }$
[$\because$ exterior angle = sum of opposite interior angles]
In figure (ii),
In $\Delta PQS$, $\angle QPS+\angle PQS+\angle PSQ={180}^{\circ }$ [sum of all the angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {55}^{\circ }+{60}^{\circ }+a={180}^{\circ }\Rightarrow {115}^{\circ }+a={180}^{\circ }\Rightarrow a={180}^{\circ }-{115}^{\circ }={65}^{\circ }$
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 $\Delta PSR,\angle PSR+\angle SPR+\angle PRS={180}^{\circ }$
[sum of all angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {115}^{\circ }+c+{40}^{\circ }={180}^{\circ }\Rightarrow {155}^{\circ }+c={180}^{\circ }\Rightarrow c={180}^{\circ }-{155}^{\circ }={25}^{\circ }$