Question

The angles of the $\Delta ABC$ and $\Delta DEF$ are given as follows;

$A=90{}^{\circ },B=30{}^{\circ },D=90{}^{\circ },$ and $E=30{}^{\circ }$

If the side $BC$ is twice the side $EF$, which of the following statement is true?

• A. $\sin B=2\sin E$
• B. $\sin E=2\sin B$
• C. $\sin B=\sin E$
• D. $\sin A=\sin B$

Answer 1

The correct option is C $\sin B=\sin E$

In $△ABC$

$\angle A={90}^{\circ }$ , $\angle B={30}^{\circ }$

We have, $\angle A+\angle B+\angle C={180}^{\circ }$

$\therefore \angle C={180}^{\circ }-{90}^{\circ }-{30}^{\circ }={60}^{\circ }$

Similarly,

In $△DEF$

$\angle D={90}^{\circ }$ , $\angle E={30}^{\circ }$

We have, $\angle D+\angle E+\angle F={180}^{\circ }$

$\therefore \angle F={180}^{\circ }-{90}^{\circ }-{30}^{\circ }={60}^{\circ }$

In $△ABC$ and $△DEF$

$\angle A=\angle D={90}^{\circ }$

$\angle B=\angle E={30}^{\circ }$

$\angle C=\angle F={60}^{\circ }$

Three corresponding pair of angles are equal. Hence both triangles are similar.

$\therefore \frac{BC}{EF}=\frac{AB}{DE}=\frac{AC}{DF}$

Consider,

$\frac{BC}{EF}=\frac{AC}{DF}$

$⟹\frac{BC}{AC}=\frac{EF}{DF}$

$⟹\sin B=\sin E$