in-a-triangle-ABC-we-have-2sinA-4cosB-6-4sinB-3cosA-1-determine-C-

Question Number 144050 by ArielVyny last updated on 21/Jun/21

i n a t r i a n g l e A B C w e h a v e

{\begin{cases} 2 s i n \hat{A} + 4 c o s \hat{B} = 6 \\ 4 s i n \hat{B} + 3 c o s \hat{A} = 1 \end{cases}

d e t e r m i n e \hat{C}

Commented by MJS_new last updated on 21/Jun/21

- 1 ⩽ \sin α ⩽ 1 \land - 1 ⩽ \cos β ⩽ 1 \Rightarrow

the only solution of the 1^{st} equation is

\sin α = 1 \land \cos β = 1 \Rightarrow

α = 90 ° \land β = 0 °

but (1) this gives no triangle and (2) this

{doesn}^{'} t fit the 2^{nd} equation

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