Question

In the adjoining figure, find the value of x.

Answer 1

Hint: Here there are two transversals OP and OQ. For a transversal passing through two parallel lines the alternate interior angles are equal. With this property we will get the angles $\mathrm{\angle }POF$ and $\mathrm{\angle }QOF$. Now, by the full angle concept we can find the angle x.
Complete step-by-step solution -

Here, we have to find the value of x.
We are given that $\mathrm{\angle }APO={45}^{\circ }$ and $\mathrm{\angle }CQO={30}^{\circ }$.
 Here we have three parallel lines AB, EF and CD. i.e.
$AB\parallel EF\parallel CD$
 There are two transversals OP and PQ.
We know that a transversal is a line that cuts two or more parallel lines at two distinct points.
 Here, OP is the transversal corresponding to the parallel lines AB and EF.
 OQ is the transversal corresponding to the parallel lines EF and CD.
Since, OP is the transversal, we can say that the alternate interior angles are equal. Therefore, we can write:
$\mathrm{\angle }APO=\mathrm{\angle }POF={45}^{\circ }$
 Similarly, for the transversal OQ the alternate interior angles are equal. Hence, we get:
$\mathrm{\angle }CQO=\mathrm{\angle }QOF={30}^{\circ }$
Here, x, $\mathrm{\angle }POF$ and $\mathrm{\angle }QOF$ form a full angle.
That is, a full angle is an angle whose measure is exactly ${360}^{\circ }$ which is a full circle.
 Hence, we will get:
$x+\mathrm{\angle }POF+\mathrm{\angle }QOF={360}^{\circ }$
 By substituting the values of $\mathrm{\angle }POF$ and $\mathrm{\angle }QOF$, we obtain:
$\begin{array}{rl}& x+{45}^{\circ }+{30}^{\circ }={360}^{\circ }\\ & x+{75}^{\circ }={360}^{\circ }\end{array}$
Now, by taking ${75}^{\circ }$ to the right side, ${75}^{\circ }$ becomes $-{75}^{\circ }$. Hence we obtain:
$\begin{array}{rl}& x={360}^{\circ }-{75}^{\circ }\\ & x={285}^{\circ }\end{array}$
 Therefore the value of $x$ is ${285}^{\circ }$.

Note: Here, you can find in other way by splitting the angle x into $\mathrm{\angle }POE$ and $\mathrm{\angle }QOE$ which you can find with the help of its alternate exterior angles $\mathrm{\angle }BPO$ and $\mathrm{\angle }DQO$, where they forms a linear pair with $\mathrm{\angle }APO$ and $\mathrm{\angle }CQO$.