Given triangle with $AB=10$, $BC=5$, $AC=12$, show that $\sin\angle BAC=\frac{3\sqrt{119}}{80}$.

Question

Given a triangle whose sides are $AB=10$, $BC=5$, and $AC=12$, how can I reach the conclusion that $\sin\angle BAC=\frac{3\sqrt{119}}{80}$?

I know that is a basic question, but I'm learning Math all by myself and I really could find an answer anywhere else. I would appreciate a detailed explanation, if you wouldn't mind.

Answer 1

Simply use the cosine rule ie $\cos c= \frac{a^2+b^2-c^2}{2ab}$ . Then use $\sin c =\sqrt{1-\cos^2 c}$.

Answer 2

You can find $\sin(A)$ by dividing the altitude dropped from $C$ by the length of $AC$

To find the altitude you divide twice the area by the base $AB=10$

To find the area use the formula $Area = \sqrt {s(s-a)(s-b)(s-c)}$ where $s$ is half of the perimeter.

Answer 3

Cosine Rule the easiest considering the other option of using Sine Rule.

$$\cos A = \cos \angle{BAC}=\frac{b^2+c^2-a^2}{2 bc}= \frac{10^2 + 17^2-5^2}{2\cdot 10 \cdot 17} = \frac{73}{80}$$

$$ \sqrt{1-\cos^2 A} = \sin A =\frac{3\sqrt{119}}{80} $$