Question

if <A and <B are acute angles such that cosA = cos B, then show that <A = < B.

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Answer 1

Step-by-step explanation:

let triangle ABC as given in figure

with angle c is equals to 90 degree

we know that cos a is equals to cos b

cos theta is equals to b upon h

therefore cos a is equals to ac upon ab

n cos b is equals to BC upon ab

AC upon a b is equals to BC upon a b

by cancelling we get AC is equals to BC

we got that triangle ABC is isosceles triangle

with ac is equals to BC

because of it is isosceles triangle base angles are same

therefore angle a is equals to angle b

Answer 2

Answer:

Hence Proved

Step-by-step explanation:

Given- <A and <B are acute angles

and cos A=cos B

therefore, <C is 90°

cos A= cos B

BC/AB =AC/AB

BC=AB

Since , <A =<B (if two sides are equal then opposite angle of the sides are also equal)

Hence Proved