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In \(\triangle ABC,\) \(CA=4\sqrt 2,\) \(CB=4\sqrt 3,\) and \(A=60^\circ.\) What is \(B\) in degrees?
 

 May 28, 2019

Best Answer 

 #1
avatar+8829 
+3

 

We are given a side-side-angle triangle, so we can use the Law of Sines to solve this.

 

\(\frac{\sin B}{b}\ =\ \frac{\sin A}{a}\\~\\ \frac{\sin B}{4\sqrt{2}}\ =\ \frac{\sin 60°}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sin 60°}{4\sqrt{3}}\cdot4\sqrt2\\~\\ \sin B\ =\ \sin 60°\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt3}{2}\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt2}{2}\\~\\ \begin{array}{lcl} B\ =\ \arcsin(\frac{\sqrt2}{2})&\quad\text{or}\quad& B\ =\ 180°-\arcsin(\frac{\sqrt2}{2})\\~\\ B\ =\ 45°&\text{or}&B\ =\ 135°\\~\\ &&\text{But}\qquad135°+60°=195°>180°\\~\\ &&\text{So }\qquad B\neq135° \end{array}\)

 

 

B  =  45°     is the only solution.

 May 28, 2019
 #1
avatar+8829 
+3
Best Answer

 

We are given a side-side-angle triangle, so we can use the Law of Sines to solve this.

 

\(\frac{\sin B}{b}\ =\ \frac{\sin A}{a}\\~\\ \frac{\sin B}{4\sqrt{2}}\ =\ \frac{\sin 60°}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sin 60°}{4\sqrt{3}}\cdot4\sqrt2\\~\\ \sin B\ =\ \sin 60°\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt3}{2}\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt2}{2}\\~\\ \begin{array}{lcl} B\ =\ \arcsin(\frac{\sqrt2}{2})&\quad\text{or}\quad& B\ =\ 180°-\arcsin(\frac{\sqrt2}{2})\\~\\ B\ =\ 45°&\text{or}&B\ =\ 135°\\~\\ &&\text{But}\qquad135°+60°=195°>180°\\~\\ &&\text{So }\qquad B\neq135° \end{array}\)

 

 

B  =  45°     is the only solution.

hectictar May 28, 2019

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