Question

If C and Z are acute angles and that cos C = cos Z prove that ∠C = ∠Z?​

Answer 1

Answer:

Step-by-step explanation:

Given C and Z are acute angles and cos C = cos Z

cos C = BCAC and cos Z = YZXZ

cos C = cos Z =>BCAC= YZXZ

Let BCYZ= ACXZ = k..... (1)

BC = k(yz) and AC = k(xz)

In right triangle ABC, using Pythagoras theorem AB = AC2−BC2−−−−−−−−−−√

AB = (k(xz)2−(k(yz)2)−−−−−−−−−−−−−−√= k(xz)2−(yz)2−−−−−−−−−−√

In right triangle xyz, using Pythagoras theorem xy = (xz)2−(yz)2−−−−−−−−−−√

ABXY = K×(XZ)2−(YZ)2√(XZ)2−(YZ)2√

Or ABXY = k ..... (1)

From 1 and 2, BCYZ = ACXZ = ABXY = k

By SSS similarity criterion, we can conclude ΔABC is similar to ΔXYZ

∴ Corresponding angles of two similar triangles will be equal.

∴ ∠C = ∠Z

Answer 2

$Given \: cos \angle C = cos \angle Z$

$We \:have \: Cos \angle C = \frac{CB}{CA}$

$and \: Cos \angle Z = \frac{ZY}{ZX}$

$Given \: cos \angle C = cos \angle Z\implies \frac{CB}{CA} = \frac{ZY}{ZX} \: --(1)$

$From \:(1) \: CB = k \times ZY$

$and \: AC = k \times ZX \: ( Where \: k\in R^{+} )$

$From \: rt. angled /: triangle \: ABC ,$

$AB = \sqrt{ AC^{2} - CB^{2}}$

$= \sqrt{ ( k \times ZX)^{2} - ( k \times ZY)^{2} }$

$= k\sqrt{ ZX^{2} - ZY^{2}}$

$From \: rt. angled \: triangle \: XYZ , XY = \sqrt{ZX^{2} - ZY^{2}}$

$So, \frac{BA}{YX} = \frac{ k\sqrt{ZX^{2} - ZY^{2}}}{\sqrt{ZX^{2} - ZY^{2}}} = k \: --(2)$

$From \: (1) \:and \:(2) \: we \:get \:$

$\frac{CB}{ZY} = \frac{CA}{ZX} = \frac{AB}{XY}$

$\therefore \triangle ABC \sim \triangle XYZ$

$( By \: S.S.S \: Similarity \: Criterion )$

$\therefore \angle C = \angle Z$

$\blue { ( Since, \: Corresponding \:angles}$

$\blue {( of \:two \: similar \: triangles \:are \: equal )}$

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