if A and b are acute angles such that cosA=cosB, then show that ANGLEA=ANGLEB.

In triangle ABC Angle A=angle B are acute angle. AngleA is a right angle. CosA = AC/AB Cos B=BC/AB CosA=CosB(given) AC/AB=BC/AB AC/BC=AB/AB(USING ALTERNANDO RULE) AC=BC IN 🔺 ABC AC=BC Angle B= angleA (opposite angle of equal sides are equal) AngleA =angle B Proved

Whoever asked this question better give up math
 

THERE IS ONE EASY METHOD TO FIND THE SOLUTION:-
Here, RHS Similarity Criterion can be used (Refer to pg. no.154 - A NOTE TO THE READER   ...NCERT TEXTBOOK).
By using this similarity criterion, the proof will become simpler.

it is very simple just open the values of cos A and cos B
cos A = AC/AB

cosB =  BC/AB

it is given cos A = cos B
then AC/AB = BC/AB

then AC = BC

then by property of angle opposite to equal side are equal

angle A = angle B

hence proved

Good answer like and i will give u the answer
 

Is there a possibility of solving by taking two triangles and then doing it similar to Example 2 on page 178.?

yes manish you are correct i did same way
 

CosA = cosB is sin(90-a)=sin(90-b) implies 90-a=90-b hence A=B......alternately construct a right triangle with A B and C as its right angled vertex ...and find cos a= cosb u will get AC=BC which implies A=B