Question

In the given figure, the measure of z is 10° less than twice the measure of y. If the difference of z and y is 20°, then the measure of (t + x) is equal to (z > y)

• A. 30°
• B. 36°
• C. 40°
• D. 45°

Answer 1

The correct option is B 36°

Let the rays AF and BE intersect at C.
In ΔABC,
∠ABC + ∠BAC = ∠ACE (Exterior angle property)
$\Rightarrow y+z=5t$
$\Rightarrow t=\frac{1}{5}(y+z)$ .....(i)
Since the measures of z is ${10}^{\circ }$ less than twice the measure of y.
$\therefore z=2y–{10}^{\circ }$
$\Rightarrow 2y–z={10}^{\circ }$ …..(ii)
Also, the difference of z and y is ${20}^{\circ }$ and z > y.
$\therefore z–y={20}^{\circ }$ …..(iii)
Adding (ii) and (iii), we get
$2y–z+z–y={10}^{\circ }+{20}^{\circ }$
$\Rightarrow y={30}^{\circ }$
Substituting $y={30}^{\circ }$ in (iii), we get
$z–{30}^{\circ }={20}^{\circ }$
$\Rightarrow z={50}^{\circ }$
Substituting $y={30}^{\circ }$ and $z={50}^{\circ }$ in (i), we get
$t=\frac{1}{5}({30}^{\circ }+{50}^{\circ })=\frac{{80}^{\circ }}{5}={16}^{\circ }$
Now, ∠BCF = ∠ACE (Vertically opposite angles)
$\Rightarrow 4x=5t$
$\Rightarrow x=\frac{5}{4}\times {16}^{\circ }={20}^{\circ }$ $(\because t={16}^{\circ })$
$\therefore x+t={20}^{\circ }+{16}^{\circ }={36}^{\circ }$
Hence, the correct answer is option (b).