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Solve right triangle ABC for all missing parts. Express angles in decimal degrees.a = 200.7 km, c= 401.5 kmRound to the nearest hundred

Solve right triangle ABC for all missing parts. Express angles in decimal degrees-example-1
User Ajeet Khan
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1 Answer

17 votes
17 votes

Using the Pythagorean Theorem we get:


c^2=a^2+b^2\text{.}

Therefore:


b^2=c^2-a^2\text{.}

Substituting a=200.7km and c=401.5km we get:


b^2=(401.5km)^2-(200.7km)^2.

Solving the above equation for b we get:


\begin{gathered} b=\sqrt[]{(401.5km)^2-(200.7km)^2} \\ =\sqrt[]{161202.25km^2-40280.79km^2} \\ =\sqrt[]{120921.76km^2}\approx347.74km\text{.} \end{gathered}

Now, from the given diagram we get that:


\begin{gathered} \cos B=(a)/(c), \\ \sin A=(a)/(c)\text{.} \end{gathered}

Substituting a=200.7km and c=401.5km we get:


\begin{gathered} \cos B=\frac{200.7\operatorname{km}}{401.5\operatorname{km}}=(2007)/(4015)\text{.} \\ \sin A=\frac{200.7\operatorname{km}}{401.5\operatorname{km}}=(2007)/(4015)\text{.} \end{gathered}

Therefore:


\begin{gathered} B=\cos ^(-1)((2007)/(4015))\approx60.00^(\circ), \\ A=\sin ^(-1)((2007)/(4015))\approx30.00^(\circ), \end{gathered}

Answer:


\begin{gathered} b=347.74\operatorname{km}, \\ B=60.00^(\circ), \\ A=30.00^(\circ) \end{gathered}

User Ashar
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