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Question Number 182896 by sciencestudent last updated on 16/Dec/22

Commented by JDamian last updated on 16/Dec/22

bigger than what?

Answered by mr W last updated on 17/Dec/22

say ∠B=20° case 1: sin ∠A=((sin 20°)/3) ⇒∠A=sin^(−1) (((sin 20°)/3))≈6.5° case 2: sin (180°−20°−A)=3 sin A tan A=(( sin 20°)/(3−cos 20°)) ⇒A=tan^(−1) ((( sin 20°)/(3−cos 20°)))≈9.4°

$${say}\:\angle{B}=\mathrm{20}° \\ $$$$\underline{{case}\:\mathrm{1}:} \\ $$$$\mathrm{sin}\:\angle{A}=\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{3}} \\ $$$$\Rightarrow\angle{A}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{3}}\right)\approx\mathrm{6}.\mathrm{5}° \\ $$$$\underline{{case}\:\mathrm{2}:} \\ $$$$\mathrm{sin}\:\left(\mathrm{180}°−\mathrm{20}°−{A}\right)=\mathrm{3}\:\mathrm{sin}\:{A} \\ $$$$\mathrm{tan}\:{A}=\frac{\:\mathrm{sin}\:\mathrm{20}°}{\mathrm{3}−\mathrm{cos}\:\mathrm{20}°} \\ $$$$\Rightarrow{A}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\:\mathrm{sin}\:\mathrm{20}°}{\mathrm{3}−\mathrm{cos}\:\mathrm{20}°}\right)\approx\mathrm{9}.\mathrm{4}° \\ $$

Commented by mr W last updated on 17/Dec/22

Commented by sciencestudent last updated on 18/Dec/22

thanks sir!

$${thanks}\:{sir}! \\ $$

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