Question

find bc. round to the nearest tenth.
(image of triangle with vertices a, b, c; ab = 7, ac = 9, angle at a is 60°)

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle, so we use the Law of Cosines. The Law of Cosines formula for a triangle with sides \(a\), \(b\), \(c\) and included angle \(C\) is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In triangle \(ABC\), we want to find \(BC\) (let's call it \(a\)), with \(AB = 7\) (let's call it \(c\)), \(AC=9\) (let's call it \(b\)), and the included angle at \(A\) is \(60^{\circ}\). So the formula becomes \(BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos(60^{\circ})\).

Step2: Substitute the Values

Substitute \(AB = 7\), \(AC = 9\), and \(\cos(60^{\circ})=\frac{1}{2}\) into the formula:
\[

$$\begin{align*} BC^{2}&=7^{2}+9^{2}-2\times7\times9\times\cos(60^{\circ})\\ &=49 + 81-2\times7\times9\times\frac{1}{2}\\ &=49 + 81-63\\ &=130 - 63\\ &= 67 \end{align*}$$

\]

Step3: Find the Length of \(BC\)

Take the square root of \(BC^{2}\) to find \(BC\):
\[
BC=\sqrt{67}\approx8.2
\]

Answer:

\(8.2\)