Question
If $ \angle B $ and $\angle Q$ are acute angles such that sin B = sin Q, then prove that $ \angle B = \angle Q$.
✓
Answer
Consider two right triangles ABC and PQR in which $ \angle B{/tex} and $\angle Q$ are the right angles.
We have,
In $\triangle ABC$
$\sin B=\frac{AC}{AB}$
and, In $\triangle PQR$
$\sin Q=\frac{PR}{PQ}$
$ \because \quad \sin B = \sin Q$
$ \Rightarrow \quad \frac { A C } { A B } = \frac { P R } { P Q }$
$ \Rightarrow \quad \frac { A C } { P R } = \frac { A B } { P Q } = k$(say) ...... (i)
$ \Rightarrow $ AC = kPR and AB = kPQ .....(ii)
Using Pythagoras theorem in triangles ABC and PQR, we obtain
$AB^2 = AC^2 + BC^2$ and $PQ^2 = PR^2 + QR^2$
$ \Rightarrow \quad B C = \sqrt { A B ^ { 2 } - A C ^ { 2 } } \text { and } Q R = \sqrt { P Q ^ { 2 } - P R ^ { 2 } }$
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { \sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \frac { \sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } }$ [ using (ii) ]
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { k \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k$...(iii)
From (i) and (iii), we get
$ \frac { A C } { P R } = \frac { A B } { P Q } = \frac { B C } { Q R }$
$ \Rightarrow \quad \Delta A C B - \Delta P R Q$ [By S.A.S similarity]
$ \therefore \quad \angle B = \angle Q$
Hence proved.
We have,
In $\triangle ABC$
$\sin B=\frac{AC}{AB}$
and, In $\triangle PQR$
$\sin Q=\frac{PR}{PQ}$
$ \because \quad \sin B = \sin Q$
$ \Rightarrow \quad \frac { A C } { A B } = \frac { P R } { P Q }$
$ \Rightarrow \quad \frac { A C } { P R } = \frac { A B } { P Q } = k$(say) ...... (i)
$ \Rightarrow $ AC = kPR and AB = kPQ .....(ii)
Using Pythagoras theorem in triangles ABC and PQR, we obtain
$AB^2 = AC^2 + BC^2$ and $PQ^2 = PR^2 + QR^2$
$ \Rightarrow \quad B C = \sqrt { A B ^ { 2 } - A C ^ { 2 } } \text { and } Q R = \sqrt { P Q ^ { 2 } - P R ^ { 2 } }$
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { \sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \frac { \sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } }$ [ using (ii) ]
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { k \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k$...(iii)
From (i) and (iii), we get
$ \frac { A C } { P R } = \frac { A B } { P Q } = \frac { B C } { Q R }$
$ \Rightarrow \quad \Delta A C B - \Delta P R Q$ [By S.A.S similarity]
$ \therefore \quad \angle B = \angle Q$
Hence proved.
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