Question

In∆ abc and ∆ xyz,if angle A and angle X are acute angles.such that cos A= cos X . Show that angle A= angle X.

Answer 1

Assuming △ABC and △XYZ are right Triangles

$\sf \bf\underline{Given} \\ \\ \sf{Cos \: A = Cos \: B} \\ \\ \sf{Cos \: A \: = \: { \frac{AB}{AC} }} \\ \\ \sf{Cos \: X \: = \: \frac{XY}{XZ} } \\ \\ \sf{ \therefore \: \: \frac{AB}{AC} = \frac{XY}{XZ} = K \: - - - - (1) } \\ \\ \sf{By \: cross \: multiplication} \\ \\ \sf{ \frac{AB}{XY} = \frac{AC}{XZ} = K \: - - - - (2)} \\ \\ \implies \sf{ \frac{BC}{YZ} = \frac{ \sqrt{ {AC}^{2} - {AB}^{2} } }{ \sqrt{ {XZ}^{2} - {xy}^{2} } }} \\ \\ \sf{ \implies \frac{BC}{ZY} = \frac{ \sqrt{ {AC}^{2} - K( {AC)}^{2} } }{ \sqrt{ {XD}^{2} - k {(XD)}^{2} } }} \: \: \: \: \: ( \because \: From \: equ \: 2)\\ \\ \sf{ \implies \frac{BC}{YZ} = \frac{ \sqrt{ {AC}^{2} - {k}^{2} {AC}^{2} } }{ \sqrt{{XZ}^{2} - {K}^{2} {XZ}^{2} } }} \\ \\ \sf{ \implies\frac{BC}{YZ} = \frac{ \sqrt{ {AC}^{2} \: { \cancel{ ({ 1 - K}^{2} )} } }}{ \sqrt{ {XZ \: }^{2} { \cancel{ {(1 - K}^{2} )} } }}} \\ \\ \sf{ \implies \frac{BC}{YZ} = \frac{ \sqrt{ {CA}^{2} } }{ \sqrt{ {XZ}^{2} } }} \\ \\ \sf{ \implies \frac{BC}{XZ} = \frac{AC}{XY} } \\ \\ \sf{ \implies \frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ} } \\ \\ \sf{By \: SSA \: \triangle ABC \thicksim \: \triangle XYZ} \\ \\ \sf{By \: properly \: of \: similarly } \\ \\ \purple{ \implies} \purple{ \underline{ \boxed{ \sf{ \angle A = \angle X}}}}$