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# help with triangle

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Find AC such that AB= 5sqrt2 and angle b= 45 and angle a=15.

Jan 9, 2021

#1
+25658
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Find AC such that AB= 5sqrt2 and angle b= 45 and angle a=15.

$$\begin{array}{|rcll|} \hline \dfrac{ \sin(45^\circ) } {AC} &=& \dfrac{ \sin\Big(180^\circ-(15^\circ+45^\circ)\Big) } {5\sqrt{2}} \\\\ \dfrac{ \sin45^\circ) } {AC} &=& \dfrac{ \sin(15^\circ+45^\circ) } {5\sqrt{2}} \\\\ \dfrac{ \sin(45^\circ) } {AC} &=& \dfrac{ \sin(60^\circ) } {5\sqrt{2}} \quad | \quad \sin(60^\circ) =\dfrac{\sqrt{3}}{2} \\\\ \dfrac{ \sin(45^\circ) } {AC} &=& \dfrac{ \sqrt{3} } {10\sqrt{2}} \quad | \quad \sin(45^\circ) =\dfrac{\sqrt{2}}{2} \\\\ \dfrac{ \sqrt{2} } {2*AC} &=& \dfrac{ \sqrt{3} } {10\sqrt{2}} \\\\ \dfrac{ \sqrt{2}\sqrt{2} } {2*AC} &=& \dfrac{ \sqrt{3} } {10} \\\\ \dfrac{ 1 } {AC} &=& \dfrac{ \sqrt{3} } {10} \\\\ AC &=& \dfrac {10}{ \sqrt{3} } * \dfrac{\sqrt{3}} {\sqrt{3}} \\\\ AC &=& \dfrac {10}{3} \sqrt{3} \\\\ \mathbf{AC} &=& \mathbf{5.7735026919} \\ \hline \end{array}$$

Jan 9, 2021