Question

If the sides of a triangle ABC are a=4,b=6 and c=8 . shown that 4cos B +CosC =2.

Answer 1

Given, A= 4,B= 6,C=8
By use of section formula.

Cos B= a^2+c^2-b^2/2ac.
= 16+64-36/2×4×8
44/64=11/16

CosC =a^2+b^2-c^2/2ab
=16+36-64/2*4*6
-12/48=-1/4
4cosB+3cosC=4*11/16-3/4
11/4-3/4 = 2.

Answer 2

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!}}}$

$<b><i><font face=Copper black size=4 color=blue>$

Here is your answer in the attachment


$\large{\red{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\underline{\underline{\underline{Hope\:it\: helps\: you}}}}}}}}}}}}}}}$