If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$,
Question.
If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$, then show that $\angle \mathrm{A}=\angle \mathrm{B}$.
If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$, then show that $\angle \mathrm{A}=\angle \mathrm{B}$.
Solution:
In figure $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles of $\triangle \mathrm{ABC}$.
Draw $C D \perp A B$
We are given that $\cos A=\cos B$
$\Rightarrow \frac{\mathbf{A D}}{\mathbf{A C}}=\frac{\mathbf{B D}}{\mathbf{B C}}$
$\Rightarrow \frac{\mathbf{A D}}{\mathbf{B D}}=\frac{\mathbf{A C}}{\mathbf{B C}}\left(\mathbf{F a c h}=\frac{\mathbf{C D}}{\mathbf{C D}}\right)$
$\Rightarrow \Delta \mathrm{ADC} \sim \Delta \mathrm{BDC} \quad(\mathrm{SSS}$ similarity criterion $) \Rightarrow \angle \mathrm{A}=\angle \mathrm{B}$
( $\because$ all the corresponding angles of two similar triangles are equal)
In figure $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles of $\triangle \mathrm{ABC}$.
Draw $C D \perp A B$
We are given that $\cos A=\cos B$
$\Rightarrow \frac{\mathbf{A D}}{\mathbf{A C}}=\frac{\mathbf{B D}}{\mathbf{B C}}$
$\Rightarrow \frac{\mathbf{A D}}{\mathbf{B D}}=\frac{\mathbf{A C}}{\mathbf{B C}}\left(\mathbf{F a c h}=\frac{\mathbf{C D}}{\mathbf{C D}}\right)$
$\Rightarrow \Delta \mathrm{ADC} \sim \Delta \mathrm{BDC} \quad(\mathrm{SSS}$ similarity criterion $) \Rightarrow \angle \mathrm{A}=\angle \mathrm{B}$
( $\because$ all the corresponding angles of two similar triangles are equal)
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.