Question
If $\angle\text{A}$ and $\angle\text{B}$ are acute angles such that $\sin\text{A}=\sin\text{B}$ then prove that $\angle\text{A}=\angle\text{B}.$
✓
Answer
Consider two right $\triangle\text{XAY}$ and WBZ such that $\sin\text{A}=\sin\text{B}$
We have,
$\sin\text{A}=\frac{\text{XY}}{\text{XA}}$ and $\sin\text{B}=\frac{\text{WZ}}{\text{WB}}$
Since $\sin\text{A}=\sin\text{B}$
$\Rightarrow\frac{\text{XY}}{\text{XA}}=\frac{\text{WZ}}{\text{WB}}$
$\Rightarrow\frac{\text{XY}}{\text{WZ}}=\frac{\text{XA}}{\text{WB}}=\text{k}(\text{say})\dots(\text{i})$
$\Rightarrow\text{XY}=\text{k}\times\text{WZ}$ and $\text{XA}=\text{k}\times\text{WB}\dots(\text{ii})$
Using Pythagoras theoram in $\triangle\text{XAY}$ and WBZ, we have
$\text{XA}^2=\text{XY}^2+\text{AY}^2$ and $\text{WB}^2=\text{WZ}^2+\text{BZ}^2$
$\Rightarrow\text{AY}=\sqrt{\text{XA}^2-\text{XY}^2}$ and $\text{BZ}=\sqrt{\text{WB}^2-\text{WZ}^2}$
$\Rightarrow\frac{\text{AY}}{\text{BZ}}=\frac{\sqrt{\text{XA}^2-\text{XY}^2}}{\sqrt{\text{WB}^2-\text{WZ}^2}}=\frac{\sqrt{\text{k}^2\text{WB}^2-\text{k}^2\text{WZ}^2}}{\sqrt{\text{WB}^2-\text{WZ}^2}}$
$\Rightarrow\frac{\text{AY}}{\text{BZ}}=\text{k}\dots(\text{iii})$
From (i), (ii) and (iii), we get
$\frac{\text{XY}}{\text{WZ}}=\frac{\text{XA}}{\text{WB}}=\frac{\text{AY}}{\text{BZ}}$
$\Rightarrow\triangle\text{XYA}\sim\triangle\text{WZB}$
$\Rightarrow\angle\text{A}=\angle\text{B}$
We have,
$\sin\text{A}=\frac{\text{XY}}{\text{XA}}$ and $\sin\text{B}=\frac{\text{WZ}}{\text{WB}}$
Since $\sin\text{A}=\sin\text{B}$
$\Rightarrow\frac{\text{XY}}{\text{XA}}=\frac{\text{WZ}}{\text{WB}}$
$\Rightarrow\frac{\text{XY}}{\text{WZ}}=\frac{\text{XA}}{\text{WB}}=\text{k}(\text{say})\dots(\text{i})$
$\Rightarrow\text{XY}=\text{k}\times\text{WZ}$ and $\text{XA}=\text{k}\times\text{WB}\dots(\text{ii})$
Using Pythagoras theoram in $\triangle\text{XAY}$ and WBZ, we have
$\text{XA}^2=\text{XY}^2+\text{AY}^2$ and $\text{WB}^2=\text{WZ}^2+\text{BZ}^2$
$\Rightarrow\text{AY}=\sqrt{\text{XA}^2-\text{XY}^2}$ and $\text{BZ}=\sqrt{\text{WB}^2-\text{WZ}^2}$
$\Rightarrow\frac{\text{AY}}{\text{BZ}}=\frac{\sqrt{\text{XA}^2-\text{XY}^2}}{\sqrt{\text{WB}^2-\text{WZ}^2}}=\frac{\sqrt{\text{k}^2\text{WB}^2-\text{k}^2\text{WZ}^2}}{\sqrt{\text{WB}^2-\text{WZ}^2}}$
$\Rightarrow\frac{\text{AY}}{\text{BZ}}=\text{k}\dots(\text{iii})$
From (i), (ii) and (iii), we get
$\frac{\text{XY}}{\text{WZ}}=\frac{\text{XA}}{\text{WB}}=\frac{\text{AY}}{\text{BZ}}$
$\Rightarrow\triangle\text{XYA}\sim\triangle\text{WZB}$
$\Rightarrow\angle\text{A}=\angle\text{B}$
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