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Question-199480




Question Number 199480 by Abdullahrussell last updated on 04/Nov/23
Commented by mr W last updated on 04/Nov/23
=((sin θ+sin 50°+1+sin 50°)/(cos θ+cos 50°+0−cos 50°))  =((sin θ+1+2 sin 50°)/(cos θ))
$$=\frac{\mathrm{sin}\:\theta+\mathrm{sin}\:\mathrm{50}°+\mathrm{1}+\mathrm{sin}\:\mathrm{50}°}{\mathrm{cos}\:\theta+\mathrm{cos}\:\mathrm{50}°+\mathrm{0}−\mathrm{cos}\:\mathrm{50}°} \\ $$$$=\frac{\mathrm{sin}\:\theta+\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\mathrm{50}°}{\mathrm{cos}\:\theta} \\ $$

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