Question

4. Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.

Answer 1

Answer:

The missing parts of the triangle are $c \approx 11.2\,km$, $A\approx 21.3^{\circ}$ and $B \approx 38.8^{\circ}$, respectively.

Step-by-step explanation:

A triangle is formed by three sides and three angles, a side and two angles are missing ($c$, $A$, $B$).

The length of the missing side is found by the Law of Cosine:

$c = \sqrt{a^{2}+b^{2}-2\cdot a \cdot b\cdot \cos C}$ (1)

($a = 4.7\,km$, $b = 8.1\,km$, $C = 119.90^{\circ}$)

$c = \sqrt{(4.7\,km)^{2}+(8.1\,km)^{2}-2\cdot (4.7\,km)\cdot (8.1\,km)\cdot \cos 119.90^{\circ}}$

$c \approx 11.2\,km$

And the missing angles can be determined by the Law of Sine:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

($a = 4.7\,km$, $b = 8.1\,km$, $c \approx 11.210\,km$, $C = 119.9^{\circ}$)

$A \approx \sin^{-1}\left(\frac{a}{c} \times \sin C \right)$

$A \approx \sin^{-1} \left(\frac{4.7\,km}{11.210\,km}\times \sin 119.9^{\circ} \right)$

$A\approx 21.3^{\circ}$

$B \approx \sin^{-1} \left(\frac{b}{c}\times \sin C \right)$

$B\approx \sin^{-1}\left(\frac{8.1\,km}{11.210\,km}\times \sin 119.9^{\circ} \right)$

$B \approx 38.8^{\circ}$

The missing parts of the triangle are $c \approx 11.2\,km$, $A\approx 21.3^{\circ}$ and $B \approx 38.8^{\circ}$, respectively.