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Question-148886




Question Number 148886 by DELETED last updated on 01/Aug/21
Answered by Rasheed.Sindhi last updated on 01/Aug/21
((AC)/(sin B))=((BC)/(sin A))  (8/(sin 60))=(6/(sin θ))  (8/((√3)/2))=(6/(sin θ))  (6/(sin θ))=((16)/( (√3)))  sin θ=((√3)/(16))×6=((3(√3))/8)  θ=sin^(−1) (((3(√3))/8))≈40.5°
$$\frac{{AC}}{\mathrm{sin}\:{B}}=\frac{{BC}}{\mathrm{sin}\:{A}} \\ $$$$\frac{\mathrm{8}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{6}}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{8}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{6}}{\mathrm{sin}\:\theta}=\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{16}}×\mathrm{6}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\right)\approx\mathrm{40}.\mathrm{5}° \\ $$

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