Question

In the figure, if l || m, $\angle 3=(x+{60}^{\circ })$ and $\angle 6=(2x+15{)}^{\circ }$, find $\angle 1$ and $\angle 8$.

Answer 1

Given, $\angle 3=(x+{60}^{\circ })$ and $\angle 6=(2x+{15}^{\circ })$

$\angle 7=\angle 3=(x+{60}^{\circ })$ ---(corresponding angles)

$\angle 6+\angle 7={180}^{\circ }$---(Angles on the straight line)

$(2x+{15}^{\circ })+(x+{60}^{\circ })={180}^{\circ }$

$3x+{75}^{\circ }={180}^{\circ }$

$3x={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

$x=\frac{105}{3}={35}^{\circ }$

$\therefore \angle 3=(x+{60}^{\circ })={95}^{\circ }$

and $\angle 6=(2x+{15}^{\circ })={85}^{\circ }$

$\angle 8=\angle 6={85}^{\circ }$ ---(vertically opposite angles)

$\angle 1=\angle 3={95}^{\circ }$ ---(vertically opposite angles)