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Find a.Round to the nearest tenth:2 cmс1501050a=a = [ ? ]cmLaw of Sines: sin A=sin Bb=sin Cсa

Find a.Round to the nearest tenth:2 cmс1501050a=a = [ ? ]cmLaw of Sines: sin A=sin-example-1
User HotFudgeSunday
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1 Answer

21 votes
21 votes

In the given triangle ABC ,

Sum of the angles of of a triangle is 180 degrees.

Therefore,


\begin{gathered} \angle A\text{ + }\angle B\text{ + }\angle C\text{ = 180} \\ \angle A\text{ + 105 + 15 = 180} \\ \angle A=\text{ 180 - 120} \\ \angle A\text{ = 60} \end{gathered}

By using sine rule,


(a)/(\sin A)\text{ = }(b)/(\sin B)

Substituting the given values in the given equation,


\begin{gathered} (a)/(\sin60)\text{ = }(2)/(\sin 105) \\ a\text{ = }(2\sin 60)/(\sin 105) \\ a\text{ = 1.7931 } \\ a\text{ }\approx\text{ 1.8 }cm \end{gathered}

Find a.Round to the nearest tenth:2 cmс1501050a=a = [ ? ]cmLaw of Sines: sin A=sin-example-1
User Zeratool
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