Question

Please help me solve this triangle so I know I’m doing it right.

Answer 1

Let the given sides be,

x=7

y=11

z=6

To find:

$\angle X,\angle Y,\angle Z$

Using the formula,

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

On substitution we get,

$\begin{gathered} \cos X=\frac{11^2+6^2-7^2}{2(11)(6)} \\ \cos X=\frac{121+36-49^{}}{132} \\ \cos X=0.81818 \\ X=\cos ^{-1}(0.81818) \\ X=35.1^{\circ} \end{gathered}$

Hence, the ange of X is,

$\angle X=35.1^{\circ}$

Next, we need to find the angle of y:

Using the formula,

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

On substitution we get,

$\begin{gathered} \cos Y=\frac{7^2+6^2-11^2}{2(7)(6)} \\ \cos Y=\frac{49+36-121^{}}{84} \\ \cos Y=-0.42857 \\ Y=\cos ^{-1}(-0.42857) \\ Y=115.4^{\circ} \end{gathered}$

Hence, the ange of Y is,

$\angle Y=115.4^{\circ}$

Next, we need to find the angle of Z:

Using the formula,

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

On substitution we get,

$\begin{gathered} \cos Z=\frac{7^2+11^2-6^2}{2(7)(11)} \\ \cos Z=\frac{49+121-36^{}}{154} \\ \cos Z=0.87012 \\ Z=\cos ^{-1}(0.87012) \\ Z=29.5^{\circ} \end{gathered}$

Hence, the ange of Z is,

$\angle Z=29.5^{\circ}$