Question

In the given figure, ABC is a triangle with $BC\parallel EF.\angle AEF={50}^o$ and $\angle FCB={70}^o$. Find $\angle A,\angle B,\angle F$.

• A. $\angle A={70}^o,\angle B={50}^0,\angle F={60}^o$
• B. $\angle A={60}^o,\angle B={50}^0,\angle F={70}^o$
• C. $\angle A={60}^o,\angle B={70}^0,\angle F={50}^o$
• D. $\angle A={50}^o,\angle B={70}^0,\angle F={60}^o$

Answer 1

Answer: (B)

$\angle A={60}^o,\angle B={50}^0,\angle F={70}^o$

Given, $EF\parallel BC$ and AB is transversal
$\therefore \angle AEF=\angle EBC$ (corresponding angles)
$\angle EBC={50}^o(As\angle AEF={50}^o)$
$\angle B={50}^o$
Also, $\angle AFE=\angle FCB$ (corresponding angles)
$\angle AFE={70}^o(As\angle FCB={70}^o)$
$\angle F={70}^o$
Now, $\angle A+\angle B+\angle C={180}^o$ (Angle sum property of triangle)
$\angle A+{50}^o+{70}^0={180}^o$
$\angle A={180}^o-{120}^o$
$={60}^o$