Question

given the triangle
, find the length of side ( x ) using the law of cosines. round your final answer to 4 decimal places.
( x = )
question help:

Explanation:

Step1: Apply the Law of Cosines formula

The Law of Cosines formula is \(x^{2}=a^{2}+b^{2}-2ab\cos C\). Here, \(a = 11\), \(b = 11\), and \(C=49^{\circ}\).
Substitute the values into the formula:
\(x^{2}=11^{2}+11^{2}-2\times11\times11\times\cos(49^{\circ})\)

Step2: Calculate each part

First, calculate \(11^{2}=121\). So \(11^{2}+11^{2}=121 + 121=242\).
Next, calculate \(2\times11\times11 = 242\).
We know that \(\cos(49^{\circ})\approx0.6561\). Then \(242\times\cos(49^{\circ})\approx242\times0.6561 = 158.7762\)
Now, \(x^{2}=242-158.7762=83.2238\)

Step3: Find \(x\)

Take the square root of \(x^{2}\): \(x=\sqrt{83.2238}\approx9.1227\)

Answer:

\(9.1227\)