Question

In ΔXYZ, XY = 6.32, YZ = 5.83, and XZ = 11.05. m∠Y = º.

Answer 1

Answer: 130.8 degrees

Explanation:

The problem can be solved by using the cosine theorem:

$a^2 = b^2 + c^2 - 2bc cos \alpha$ (1)

where

a,b,c are the lengths of the three sides of the triangle

$\alpha$ is the angle between b and c

In this problem, we can identify a,b,c, with:

a = XZ = 11.05

b = XY = 6.32

c = YZ = 5.83

So, the angle $\alpha$ corresponds to the angle m∠Y. Re-arranging eq.(1), we find

$cos \alpha = \frac{b^2+c^2-a^2}{bc}=\frac{(6.32)^2+(5.83)^2-(11.05)^2}{2(5.83)(6.32)}=-0.654$

So, the angle is

$\alpha=cos^{-1}(-0.654)=130.8^{\circ}$

Answer 2

Answer:

130.8 degrees