Question

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

Answer 1

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:

$\begin{gathered} \cos B=\frac{a}{c}, \\ \sin A=\frac{a}{c}\text{.} \end{gathered}$

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

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

Therefore:

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

Answer:

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