In triangle ABC
Angle A=angle B are acute angle.
AngleA is a right angle.
CosA = AC/AB
AC/BC=AB/AB(USING ALTERNANDO RULE)
IN 🔺 ABC
Angle B= angleA (opposite angle of equal sides
AngleA =angle B
Shashi Nandi answered this
Whoever asked this question better give up math
Manish Mahasubh answered this
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.
Parav answered this
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
Samyak answered this
Good answer like and i will give u the answer
Ketul answered this
Is there a possibility of solving by taking two triangles and then doing it similar to Example 2 on page 178.?
W I T T Y answered this
yes manish you are correct i did same way
Kanishk answered this
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