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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

#1
+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
+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.

hectictar May 28, 2019