If angleA and angleB are acute angles such that cosA=cosB then show that angleA=angleB
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Answered by
11
In a triangle
Cos A = cos B
AC/AB= BC/AB
⇒AC=BC
⇒Angle A and Angle B
∠A = ∠B
Cos A = cos B
AC/AB= BC/AB
⇒AC=BC
⇒Angle A and Angle B
∠A = ∠B
Answered by
7
It's given that cosA=cosB. eq. (1)
in ACB angle C= 90 degree
cosA=AC/AB
cosB=BC/AB
from eq. (1)
cosA=cosB
AC/AB =BC/AB
AC=BC
angleB=angleA (angles opposite to equal sides are equal)
in ACB angle C= 90 degree
cosA=AC/AB
cosB=BC/AB
from eq. (1)
cosA=cosB
AC/AB =BC/AB
AC=BC
angleB=angleA (angles opposite to equal sides are equal)
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