Question

pls answer correctly ​

Answer 1

Step-by-step explanation:

According to your question both angle A and B are acute,

and CosA = CosB (given)

we have to prove that angleA = angleB

$\cos = \frac{base}{hypotenuse}$

According to cos(A) ,

AB = hypotenuse

CB = prependicular

AC = base

$\cos(a) = \frac{base}{hypotenuse} = \frac{ac}{ab}$

Now , According to cos(B)

AB = hypotenuse

BC = base

CA = prependicular

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

and it's given that

$\cos(a) = \cos(b) \\ \\ = \geqslant \frac{ac}{ab} = \frac{bc}{ab} \\ \\ = \geqslant cross \: multiplying \\ \\ = \geqslant ac \times ab = ab \times bc \\ \\ = \geqslant ab \: will \: be \: cancelled \\ \\ = \geqslant ac = bc$

Here AC = angleB (angle opposite to equal sides are equal)

BC = angleA

AC = BC

angleB = angleA. or. angleA = angleB

Hence , Proved.

Answer 2

Answer:

According to your question both angle A and B are acute,

and CosA = CosB (given)

we have to prove that angleA = angleB

\cos = \frac{base}{hypotenuse}cos=

hypotenuse

base

According to cos(A) ,

AB = hypotenuse

CB = prependicular

AC = base

\cos(a) = \frac{base}{hypotenuse} = \frac{ac}{ab}cos(a)=

hypotenuse

base

=

ab

ac

Now , According to cos(B)

AB = hypotenuse

BC = base

CA = prependicular

\cos(b) = \frac{base}{hypotenuse} = \frac{bc}{ab}cos(b)=

hypotenuse

base

=

ab

bc

and it's given that

\begin{lgathered}\cos(a) = \cos(b) \\ \\ = \geqslant \frac{ac}{ab} = \frac{bc}{ab} \\ \\ = \geqslant cross \: multiplying \\ \\ = \geqslant ac \times ab = ab \times bc \\ \\ = \geqslant ab \: will \: be \: cancelled \\ \\ = \geqslant ac = bc\end{lgathered}

cos(a)=cos(b)

=⩾ abac

= abbc

=⩾crossmultiplying

=⩾ac×ab=ab×bc

=⩾abwillbecancelled

=⩾ac=bc

Here AC = angleB (angle opposite to equal sides are equal)

BC = angleA

AC = BC

angleB = angleA. or. angleA = angleB

Hence , Proved.