Question

a parallelogram has side lengths of 4 and 6 and an angle of measure 85°
what is x, the length of the diagonal, to the nearest whole number?
law of cosines: $a^2 = b^2 + c^2 - 2bc\cos(a)$

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\). In triangle \(PQO\), let \(b = 4\), \(c = 6\), and \(A=65^{\circ}\). Then \(x^{2}=4^{2}+6^{2}-2\times4\times6\times\cos(65^{\circ})\).

Step2: Calculate each term

First, \(4^{2}=16\), \(6^{2}=36\), and \(2\times4\times6 = 48\). \(\cos(65^{\circ})\approx0.4226\). So \(x^{2}=16 + 36-48\times0.4226\).

Step3: Simplify the expression

\(x^{2}=52-20.2848=31.7152\).

Step4: Find \(x\)

Take the square - root of \(x^{2}\), \(x=\sqrt{31.7152}\approx5.63\). Rounding to the nearest whole number, \(x\approx6\).

Answer:

B. 6