Question

In $△ABC$ and $△XYZ$, if $\angle A$ and $\angle X$ are acute angles such that $\cos A=\cos x$ then show that $\angle A=\angle X$.

Answer 1

GIven,

for $△ABC$ and $△XYZ$

$\angle A$ and $\angle X$ are acute trianlge

Where $\cos A=\cos X$

Also given, to show that $\angle A=\angle X$

$\because \cos =\frac{lengthofadjside}{lengthofhypotenuse}$

$\Rightarrow \frac{AB}{AC}=\frac{XY}{XZ}$

Let $\frac{AB}{AC}=\frac{XY}{XZ}=k$ [where $k$ is a constant]

So we get, $\frac{AB}{XY}=\frac{AC}{XZ}=k\to eq\left(1\right)$

taking then separately we get,

$\frac{AB}{XY}=k\Rightarrow AB=k\times y$, parallel $\frac{AC}{XZ}=k\Rightarrow AC=kXZ$

now let us consider both the triangles opposite sides with their distance $\sqrt{x_2^2-x_1^2}=d$ we get,

$\frac{BC}{ZY}=\frac{\sqrt{A{C}^2-A{B}^2}}{\sqrt{X{Z}^2-X{Y}^2}}$ [from eqb (1) & eqn (2) we get $AB=KXY,AC=KXZ$]

by substituting the values we get

$\frac{BC}{ZY}=k\frac{\sqrt{X{Z}^2-X{Y}^2}}{\sqrt{X{Z}^2-X{Y}^2}}\Rightarrow \frac{\sqrt{k^2}\left[\sqrt{X{Z}^2-X{Y}^2}\right]}{\sqrt{X{Z}^2-X{Y}^2}}$

$\Rightarrow \frac{BC}{ZY}=k\frac{\sqrt{X{Z}^2-X{Y}^2}}{\sqrt{X{Z}^2-X{Y}^2}}\iff \frac{BC}{XY}=k$

[Let us now substitute $k$ in eq $\left(1\right)$]

$\Rightarrow \frac{AB}{XY}=\frac{AC}{XZ}=\frac{BC}{ZY}$

$\therefore$ By $SSS$ similarity we get $△ABC\sim △XYZ$

$\therefore$ By property of similarity $\angle A=\angle X$