Question 1106077
The problem seems designed as an example of law of cosines.
Right triangles are easier to solve, but if you cannot find a right angle,
your tools are law of sines and law of cosines.
Law of sines applies easily when you have the measures of an angle and the opposite side. If you only have 3 sides, you use law of cosines.
Law of cosines is given as a formula like
{{{a^2=b^2+c^2-2*b*c*cos(A)}}} that turns into {{{a^2=b^2+c^2}}} when {{{A=90^o}}} .
With that formula and the lengths of 3 sides, you call the angle you are looking for angle {{{A}}} and solve for A.
If you are supposed to write the formula, write it like the instructor or the book prefers it.
It could be {{{c^2=a^2+b^2-2*a*b*cos(C)}}} , and you would be looking for angle {{{C}}} .


1. Triangle VUW has side lengths {{{9}}} , {{{9}}} , and {{{5}}} .
The shorter side, with length {{{5}}} is opposite vertex U,
so angle {{{U}}} is the smallest angle,
and {{{5}}} is the side length that is squared on one the left side of the equal sign in the law of cosines formula.
{{{5^2=9^2+9^2-2*9*9*cos(U)}}}
{{{25=81+81-162cos(U)}}}
{{{25=162-162cos(U)}}}
{{{-137=-162cos(U)}}}
{{{cos(U)=137/162}}}
{{{cos(U)=0.84568}}} (rounded)
{{{U=32.26^o}}} (rounded)
 
2. The side lengths of triangle UWX are {{{9}}} , {{{8}}} , and {{{7}}} .
The longer side, with length {{{9}}} is opposite vertex X,
so angle {{{X}}} is the largest angle,
and {{{9}}} is the side length that is squared on one the left side of the equal sign in the law of cosines formula.
{{{9^2=8^2+7^2-2*8*7*cos(X)}}}
{{{81=64+49-112cos(X)}}}
{{{81-113-112cos(X)}}}
{{{-32=-112cos(X)}}}
{{{cos(X)=32/112}}}
{{{cos(X)=0.2857}}} (rounded)
{{{X=73.40^o}}} (rounded)