Question

Good Morning Mates !
If < A and < B are acute angles such that Cos A = cos B ,then show that < A = < B
Question from Chapter ' Trigonometry ' Class 10th .​

Answer 1

Answer:

According to the question:-

Let cosA=

Hypotenuse

sideadjacentA

AB

AC

Similarly,

cosB=

Hypotenuse

sideadjacentB

=

AB

BC

Given that

cosA=cosB

AB

AC

=

AB

BC

AC=BC

In triangle,

angles opposite equal side are equal

∠B=∠A

Step-by-step explanation:

Mark me as brainliest please.....

Answer 2

Answer:

$\cos(a) = \frac{b}{h} = \frac{ac}{ab}$

$\cos(b) = \frac{b}{h} = \frac{bc}{ab}$

$\bold \blue{by \: the \: condition}$

$\cos(a) = \cos(b)$

$= \frac{ac}{ab} = \frac{bc}{ab}$

$ac \: = \: bc$

$\bold \blue{now \: in \: triangle \: abc}$

$ac \: = bc \: \: (proved)$

$< b \: = \: < a \: ( < \: opposite \: to \: the \: equal \: sides)$

$or \: < a \: = \: < b$