Question

question / (1 point)
what is the value of c to the nearest tenth of a metre?
7 m
8 m
50°
6.4 m
41.0 m
113.9 m
10.7 m

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula for finding side $c$ in a triangle with sides $a$, $b$ and included - angle $C$ is $c^{2}=a^{2}+b^{2}-2ab\cos C$. Here, $a = 7$, $b = 8$, and $C = 50^{\circ}$.
$c^{2}=7^{2}+8^{2}-2\times7\times8\times\cos(50^{\circ})$

Step2: Calculate each term

First, calculate $7^{2}=49$, $8^{2}=64$, and $2\times7\times8 = 112$. Then, $\cos(50^{\circ})\approx0.6428$.
$c^{2}=49 + 64-112\times0.6428$
$c^{2}=49 + 64 - 71.9936$
$c^{2}=113 - 71.9936$
$c^{2}=41.0064$

Step3: Find the value of $c$

Take the square - root of both sides: $c=\sqrt{41.0064}\approx6.4$ (rounded to the nearest tenth).

Answer:

$6.4$ m