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Please look at photo for accurate description and remember to round to the nearest 10th

Please look at photo for accurate description and remember to round to the nearest-example-1
User Oleg Golovanov
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1 Answer

14 votes
14 votes

There are two possible solutions for the triangle. For both, we can use the Law of sines.


(\sin (A))/(a)=(\sin (B))/(b)=(\sin (C))/(c)\Rightarrow\text{ Law of sines}

Then, we have:


\begin{gathered} a=97 \\ b=52 \\ B=28\text{\degree} \end{gathered}

First possible solution

We find the angle A:


\begin{gathered} (\sin(A))/(a)=(\sin(B))/(b) \\ (\sin(A))/(97)=(\sin(28))/(52) \\ \text{ Multiply by 97 from both sides of the equation} \\ (\sin(A))/(97)\cdot97=(\sin(28))/(52)\cdot97 \\ \sin (A)=(97\sin(28))/(52) \\ \sin (A)=0.8757 \\ \text{ We apply the inverse function }\sin ^(-1)(x)\text{ from both sides of the equation} \\ \sin ^(-1)(\sin (A))=\sin ^(-1)(0.8757) \\ A\approx61.1\text{\degree} \end{gathered}

We find the angle C using the angle sum theorem, which says that the sum of measures of interior angles of a triangle is 180°.


\begin{gathered} A+B+C=180\text{\degree} \\ 61.1\text{\degree}+28\text{\degree}+C=180\text{\degree} \\ 89.1\text{\degree}+C=180\text{\degree} \\ \text{ Subtract 89.1\degree from both sides} \\ 89.1\text{\degree}+C-89.1\text{\degree}=180\text{\degree}-89.1\text{\degree} \\ C=90.1\text{\degree} \end{gathered}

We find the side c:


\begin{gathered} (\sin(B))/(b)=(\sin(C))/(c) \\ \frac{\sin(28\text{\degree})}{52}=\frac{\sin(90.1\text{\degree})}{c} \\ \text{ Apply cross product} \\ \sin (28\text{\degree})\cdot c=\sin (90.1\text{\degree})\cdot52 \\ \text{ Divide by }\sin (28\text{\degree})\text{ from both sides} \\ \frac{\sin (28\text{\degree})\cdot c}{\sin (28\text{\degree})}=\frac{\sin (90.1\text{\degree})\cdot52}{\sin (28\text{\degree})} \\ c\approx110.8\text{\degree} \end{gathered}

Second possible solution

We find the angle A:


\begin{gathered} (\sin(A))/(a)=(\sin(B))/(b) \\ (\sin(A))/(97)=(\sin(28))/(52) \\ \text{ Multiply by 97 from both sides of the equation} \\ (\sin(A))/(97)\cdot97=(\sin(28))/(52)\cdot97 \\ \sin (A)=(97\sin(28))/(52) \\ \sin (A)=0.8757 \\ \text{ We apply the inverse function }\sin ^(-1)(x)\text{ from both sides of the equation and we subtract this value from 180\degree} \\ 180\text{\degree}-\sin ^(-1)(\sin (A))=180\text{\degree}-\sin ^(-1)(0.8757) \\ A\approx180\text{\degree}-61.1\text{\degree} \\ A\approx118.9\text{\degree} \end{gathered}

We find the angle C using the angle sum theorem:


\begin{gathered} A+B+C=180\text{\degree} \\ 118.9\text{\degree}+28\text{\degree}+C\approx180\text{\degree} \\ 146.9\text{\degree}+C\approx180\text{\degree} \\ \text{ Subtract 146.9\degree{}from both sides} \\ 146.9\text{\degree}+C-146.9\text{\degree}\approx180\text{\degree}-146.9\text{\degree} \\ C\approx33.1\text{\degree} \end{gathered}

We find the side c:


\begin{gathered} (\sin(B))/(b)=(\sin(C))/(c) \\ \frac{\sin(28\text{\degree})}{52}=\frac{\sin(90.1\text{\degree})}{c} \\ \text{ Apply cross product} \\ \sin (28\text{\degree})\cdot c=\sin (90.1\text{\degree})\cdot52 \\ \text{ Divide by }\sin (28\text{\degree})\text{ from both sides} \\ \frac{\sin (28\text{\degree})\cdot c}{\sin (28\text{\degree})}=\frac{\sin (90.1\text{\degree})\cdot52}{\sin (28\text{\degree})} \\ c\approx110.8\text{\degree} \end{gathered}

Please look at photo for accurate description and remember to round to the nearest-example-1
User Mike Silvis
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