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Question

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


Solution

GIven, 
for $$\triangle ABC$$ and $$\triangle XYZ$$
$$\angle A$$ and $$\angle X$$ are acute trianlge
Where $$\cos A=\cos X$$
Also given, to show that $$\angle A=\angle X$$
$$\because\cos=\dfrac{length\ of\ adj\ side}{length\ of\ hypotenuse}$$
$$\Rightarrow \dfrac{AB}{AC}=\dfrac{XY}{XZ}$$
Let $$\dfrac{AB}{AC}=\dfrac{XY}{XZ}=k$$   [where $$k$$ is a constant]
So we get, $$\dfrac{AB}{XY}=\dfrac{AC}{XZ}=k\rightarrow eq (1)$$
taking then separately we get, 
$$\dfrac{AB}{XY}=k\Rightarrow AB=k\times y$$, parallel $$\dfrac{AC}{XZ}=k\Rightarrow AC=kXZ$$
now let us consider both the triangles opposite sides with their distance $$\sqrt{x_{2}^{2}-x_{1}^{2}}=d$$ we get, 
$$\dfrac{BC}{ZY}=\dfrac{\sqrt{AC^{2}-AB^{2}}}{\sqrt{XZ^{2}-XY^{2}}}$$      [from eqb (1) & eqn (2) we get $$AB=KXY, AC=KXZ$$]
by substituting the values we get
$$\dfrac{BC}{ZY}=k\dfrac{\sqrt{XZ^{2}-XY^{2}}}{\sqrt{XZ^{2}-XY^{2}}}\Rightarrow \dfrac{\sqrt{k^{2}}[\sqrt{XZ^{2}-XY^{2}}]}{\sqrt{XZ^{2}-XY^{2}}}$$
$$\Rightarrow \dfrac{BC}{ZY}=k\dfrac{\sqrt{XZ^{2}-XY^{2}}}{\sqrt{XZ^{2}-XY^{2}}}\Leftrightarrow \dfrac{BC}{XY}=k$$
[Let us now substitute $$k$$ in eq $$(1)$$]
$$\Rightarrow \dfrac{AB}{XY}=\dfrac{AC}{XZ}=\dfrac{BC}{ZY}$$
$$\therefore$$ By $$SSS$$ similarity we get $$\triangle ABC\sim \triangle XYZ$$
$$\therefore$$ By property of similarity $$\angle A =\angle X$$

1426843_1052452_ans_0bec9b9b8ac942b48d33c5615d38d5ac.png

Mathematics

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