Question

problem

1. determine the measures of ∠a and ∠c to the nearest tenth of a degree.

Explanation:

Step1: Find $\sin C$

In right - triangle $BDC$, $\sin C=\frac{BD}{BC}$. Given $BD = 10$ and $BC$ is the hypotenuse of $\triangle BDC$. First, find $\sin C=\frac{10}{\sqrt{10^{2}+6^{2}}}=\frac{10}{\sqrt{100 + 36}}=\frac{10}{\sqrt{136}}=\frac{10}{2\sqrt{34}}=\frac{5}{\sqrt{34}}$. Then $C=\sin^{- 1}(\frac{5}{\sqrt{34}})\approx59.0^{\circ}$.

Step2: Find $\angle A$

In right - triangle $ABC$, $\angle A+\angle C = 90^{\circ}$. So $A=90^{\circ}-C$. Since $C\approx59.0^{\circ}$, then $A\approx90^{\circ}-59.0^{\circ}=31.0^{\circ}$.

Answer:

$\angle A\approx31.0^{\circ}$, $\angle C\approx59.0^{\circ}$