Question

a parallelogram has side lengths of 13 and 17 and an angle that measures 64°. what is x, the length of the diagonal, to the nearest whole number? law of cosines: $a^{2}=b^{2}+c^{2}-2bccos(a)$

Explanation:

Step1: Identify values for law of cosines

Let $b = 13$, $c = 17$, and $A=64^{\circ}$. The diagonal $x$ is $a$ in the law - of - cosines formula $a^{2}=b^{2}+c^{2}-2bc\cos(A)$.

Step2: Substitute values into formula

$x^{2}=13^{2}+17^{2}-2\times13\times17\times\cos(64^{\circ})$
$x^{2}=169 + 289-442\times\cos(64^{\circ})$
We know that $\cos(64^{\circ})\approx0.4384$.
$x^{2}=169 + 289-442\times0.4384$
$x^{2}=169 + 289 - 193.8728$
$x^{2}=458-193.8728$
$x^{2}=264.1272$

Step3: Solve for $x$

$x=\sqrt{264.1272}\approx16.25$
Rounding to the nearest whole number, $x\approx16$.

Answer:

16