Question

a triangle has sides a = 2 and b = 5 and angle c = 40°. find the length of side c.
the length of side c is \boxed{}.
(round to three decimal places as needed.)

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\) is given by \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\).
Here, \(a = 2\), \(b=5\) and \(C = 40^{\circ}\). First, we need to find the value of \(\cos(40^{\circ})\). We know that \(\cos(40^{\circ})\approx0.7660\) (using a calculator).

Step2: Substitute the values into the formula

Substitute \(a = 2\), \(b = 5\) and \(\cos(C)=\cos(40^{\circ})\approx0.7660\) into the formula \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\):
\[

$$\begin{align*} c^{2}&=2^{2}+5^{2}-2\times2\times5\times\cos(40^{\circ})\\ &=4 + 25- 20\times0.7660\\ &=29-15.32\\ &=13.68 \end{align*}$$

\]

Step3: Find the value of \(c\)

To find \(c\), we take the square root of \(c^{2}\). So \(c=\sqrt{13.68}\approx3.699\) (rounded to three decimal places).

Answer:

\(3.699\)