Question

Match the following columns:

Column I	Column II
(a) An angle is 10° more than its complement. The measure of the angle is ...... .	(p) 160°
(b) If in ∆ABC, ∠A = 65° and ∠B − ∠C = 25°, then ∠B = ......	(q) 50°
(c) In ∆ABC, ∠A = 40° and ∠B = ∠C. If EF || BC, then ∠EFC = ...... .	(r) 70°
(d) If the angles around a point are 2x°, 3x°, 5x° and 40°, the measure of the largest angle is ......	(s) 110°

(a) ......
(b) ......
(c) ......
(d) ......

Answer 1

(a)
Let the required angle be $x°$.
Then, measure of its complement = $\left(90-x\right)°$.
$$\begin{gathered} \therefore x=\left(90-x\right)+10 \\ \Rightarrow 2x=100° \\ \Rightarrow x=50° \end{gathered}$$
Hence, (a)$\Rightarrow$(q).

(b)
We have:
$$\begin{gathered} \angle A+\angle B+\angle C=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 65°+\angle B+\angle C=180° \\ \Rightarrow \angle B+\angle C=115°...\left(i\right) \\ \text{Also,}\angle B-\angle C=25°...\left(ii\right)\left[\text{Given}\right] \end{gathered}$$
$$\begin{gathered} \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(ii\right),\text{we get:} \\ 2\angle B=140° \\ \Rightarrow \angle B=70° \end{gathered}$$
Hence, (b)$\Rightarrow$(r).

(c)
In $∆ABC$, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 40°+\angle B+\angle B=180°\left[\because \angle B=\angle C\right] \\ \Rightarrow 40°+2\angle B=180° \\ \Rightarrow 2\angle B=140° \\ \Rightarrow \angle B=70° \end{gathered}$$
$EF\parallel BC\text{and}AEB$is the transversal.
$\therefore \angle ABC=\angle AEF=70°\left[\text{Corresponding Angles}\right]$
Now, side AF of triangle AEF is produced to C.
$$\begin{gathered} \therefore \angle EFC=\angle EAF+\angle AEF \\ \Rightarrow \angle EFC=40°+70° \\ \Rightarrow \angle EFC=110° \end{gathered}$$
Hence, (c)$\Rightarrow$(s).

(d)
We know that the sum of the angles around a point is $360°$.
$$\begin{gathered} \therefore 2x+3x+5x+40°=360° \\ \Rightarrow 10x=320° \\ \Rightarrow x=32° \end{gathered}$$
Hence, the angles are
$2\left(32\right)°=64°,3\left(32\right)°=96°,5\left(32\right)°=160°\text{and}40°$
Therefore, the largest angle is $160°$.
Hence, (d)$\Rightarrow$(p).