Question

Determine the value of x in the given figure.

A) $55^\circ$

B) $65^\circ$

C) $68^\circ$

D) $53^\circ$

E) None of the above

Answer 1

Hint: In such a type of question first find the relation between angles. Draw the figure as per question Assume the unknown angles as $\alpha$  and $\beta$. In a triangle sum of all interior angles is $180^\circ$. That is in $\Delta ABC$ we have $\angle A + \angle B + \angle C= 180^\circ$, Write the equations of related angles and solve it in order to get the solution of the given question.

Complete step-by-step solution:

In the above figure let us assume that

$\angle BCA = {\alpha}^\circ$

And 

$\angle ABC = {\beta}^\circ$

As we see from question $ACQ$ is a straight line so we can say that $\angle QCB$ and $\angle ACB$ are supplementary angles.

So, as we know that sum of supplementary angles is $180^\circ$ so we can write

$\angle QCB + \angle ACB = 180^\circ $

So further we can write

$\angle ACB + \angle CBA = 180^\circ$

$2x^\circ + \alpha ^\circ=180^\circ$

$ \Rightarrow \alpha^\circ=180^\circ-2x^\circ$      ............................ $(1)$

Also, $\angle ABC+ \angle PBC = 180^\circ$ as both angles are supplementary, so we can write sum of supplementary angles is equals to $180^\circ$

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

$ \Rightarrow \beta^\circ = 180^\circ - {(2x-15)}^\circ$   ............................$(2)$

Now in $\Delta ABC$ we know that sum of all angles is equal to $180^\circ$

Hence, we can write

$\angle A + \angle B + \angle C = 180^\circ \Rightarrow x^\circ+ \beta ^\circ + \alpha ^\circ = 180^\circ$

Here we have write the value of $\alpha ^\circ,\,\beta ^\circ$ from equation $(2)$ and $(1)$

So, we can write here sum of all angle of the $\Delta ABC$ is equal to $180^\circ$

Hence,

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

Subtracting both side $180^\circ$ we can write

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

$ \Rightarrow x^\circ - 2x^\circ + 15^\circ + 180 ^\circ - 2x^ \circ= 0^\circ$

$-3x^\circ+195^\circ=0^\circ$

So, on transposing we can write $-3x^\circ$ on right hand side its sign will be changed so we can write

$3x^\circ=195^\circ$

Dividing both side by $3$ we get the value of $x^\circ$

$\dfrac{3x^\circ}{3} =\dfrac{195^\circ}{3}$

$\Rightarrow x=65^\circ$

Hence, we get the value of $\angle A=65^\circ$.

Therefore, option (B) is correct.

Note: A triangle is the simplest possible polygon. It is a two-dimensional shape with three sides and three interior angles. 

Two angles are called supplementary angles if the sum of both angles is $180^\circ$. If two angles made of a straight-line sum of angles is equal to $180^\circ$.