Question

find the value of a b and c

Answer 1

hope it helps!;!!;!;

Answer 2

$\pink{\underline{\huge{\star\: Solution \: \star}}}$

$\green{\underline{Equation}}$

$sin(A+B+C) = 1\\ \therefore sin \:90\degree = 1 \\ \implies sin(A+B+C) = sin 90\degree \\ \implies (A+B+C) = 90\degree ...( so A,B,C >90\degree) \\ tan(A-B) =\frac{1}{\sqrt3} \\ \therefore tan \:30\degree =\frac{1}{\sqrt3} \\ \implies tan(A-B) = tan \:30\degree \\ \implies A-B = 30\degree \\ sec(A+C) = 2 \\ A+C = sin^- 2 \\\implies A+C = 60\degree \\\therefore sec \:60 = 2$

$A+B+C = 90\degree \\ (A+C)+B = 90\\ 60\degree +B = 90\degree \\ B = 90\degree - 60\degree \\ B = 30\degree\\\\ A-B = 30\degree \\ A - 30 = 30\\ A = 30+30\\ A = 60 \\\\ A + B+ C = 90\degree \\ 30+60+C = 90\degree \\ C = 90\degree -(90\degree)\\ C = 0\degree$

$\red{\underline{Answer}}$

$\pink{\boxed{\angle{A} = 60\degree , \: \angle{B} = 30\degree , \: \angle{C} = 0\degree}}$