# In triangle XYZ if x=1, y=2, z=sqrt5, how do you find the exact value of cosZ?

Jan 14, 2017

$\cos Z = 0$

#### Explanation:

As $x = 1$, $y = 2$ and $z = \sqrt{5}$,

it is apparent that the square on the largest side $z = \sqrt{5}$ is $5$ and is equal to sum of the squares on other two sides of the triangle as $5 = {1}^{2} + {2}^{2}$

Hence $z$ is hypotenuse and $m \angle Z = \frac{\pi}{2}$

and $\cos Z = 0$

Jan 14, 2017

Applying cosine law for triangle we can write

$\cos Z = \frac{{x}^{2} + {y}^{2} - {z}^{2}}{2 x y}$

$\implies \cos Z = \frac{{1}^{2} + {2}^{2} - {\left(\sqrt{5}\right)}^{2}}{2 \cdot 1 \cdot 2} = \frac{5 - 5}{4} = 0$