Would the answer be 12.5

Mathematics

1 answer:

matrenka [14]3 years ago

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Yes hope this helps man!!

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In many cases the Law of Sines works perfectly well and returns the correct missing values in a non-right triangle. However, in

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$\frac{\sin B}{b}=\frac{\sin A}{a} \\ \\ \sin B=\frac{b \sin A}{a} \\ \\ \sin B=\frac{13 \sin 29^{\circ}}{7.2} \\ \\ \boxed{m\angle B =\arcsin \left(\frac{65\sin 29^{\circ}{36} \right), 180^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$

Considering the diagram. $\angle B$ is shown to be acute, so $\boxed{m\angle B=\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$ .

Angles in a triangle add to 180°, so $m\angle C= 151^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)$ .

Using the law of cosines,

$AB=\sqrt{13^2 + 7.2^2-2(13)(7.2)\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)} \\ \\ \boxed{AB=\sqrt{220.84-187.2\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)}}$