Question

if ΔA and ΔB are acute angles such that cos A =cos B than prove that ΔA = ΔB
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Answer 1

$\huge\boxed{\underline{{\red{Answer}}:-}}$

$\small\bf{Cos\:A=}$$\bf\frac{Sides\:of\:adjacent\:A}{Hypotenuse}$

$\rightarrow$$\bf\frac{AC}{AB}$

$\therefore$$\large\bf{Similarly,}$

$\small\bf{Cos\:B=}$$\bf\frac{Sides\:of\:adjacent\:B}{Hypotenuse}$

$\rightarrow$$\bf\frac{BC}{AB}$

______________________________________

$\therefore$$\small\bf{CosA=CosB}$

$\rightarrow$$\bf\frac{AC}{AB}$${=}$$\bf\frac{BC}{AB}$

$\therefore$$\small\bf{AC=BC}$

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ ɪɴ ᴀ ᴛʀɪᴀɴɢʟᴇ,ᴀɴɢʟᴇ ᴏᴘᴘᴏsɪᴛᴇ ᴇϙᴜᴀʟ sɪᴅᴇs ᴀʀᴇ ᴇϙᴜᴀʟ.

$\small\angle{B}$$\small{=}$$\small\angle{A}$

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ

Answer 2

let's assume the triangle ABC in which angle C=90°

Given that the angles A and B are acute angles such that cosA=cosB

As per the angles, taken the cos ratio is written as ,

AC/AB=BC/AB

=>AC=BC

so, angle A=angleB{Angles opposite to equal sides are equal}