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

$$inatriangleABCwehave$$$$\{\begin{array}{l}2sin\hat{A}+4cos\hat{B}=6\\ 4sin\hat{B}+3cos\hat{A}=1\end{array}$$$$determine\hat{C}$$

Commented by MJS_new last updated on 21/Jun/21

$$-1⩽\sin \alpha ⩽1\wedge -1⩽\cos \beta ⩽1\Rightarrow$$$$\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{of}\mathrm{the}{1}^{\mathrm{st}}\mathrm{equation}\mathrm{is}$$$$\sin \alpha =1\wedge \cos \beta =1\Rightarrow$$$$\alpha =90°\wedge \beta =0°$$$$\mathrm{but}\left(1\right)\mathrm{this}\mathrm{gives}\mathrm{no}\mathrm{triangle}\mathrm{and}\left(2\right)\mathrm{this}$$$${\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{fit}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{equation}$$

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