Answer :
WX=3.6 units
m<W=[tex]{56.3}^{0} [/tex]
m<X=[tex]{33.7}^{0} [/tex]
Step-by-step explanation:
We can calculate WX
using the Pythagorean theorem.
From the theorem !
[tex] { |WX|}^{2} ={ |WY|}^{2}+{|XY|}^{2} [/tex]
From the question,WY=2 units and
XY=3 units.
Thus,by substitution we obtain
[tex] { |WX|}^{2} ={2}^{2}+{3}^{2} [/tex]
We then simplify the right hand side of the equation
[tex]{ |WX|}^{2} =4 + 9[/tex]
[tex] \implies { |WX|}^{2} =13[/tex]
we then take the positive square root of both sides
[tex]\implies \sqrt|WX| = \sqrt{13} [/tex]
[tex]\implies |WX| = 3.6 [/tex]
We can use the tan ratio to find for the angles.
[tex] \tan(\theta) = \frac{opposite}{adjacent }[/tex]
To continue,
Let m<W=a
This implies that
[tex] \tan(a) = \frac{3}{2} [/tex]
[tex] \implies(a) = { \tan }^{ - 1} (\frac{3}{2})[/tex]
[tex]\implies(a) = {56.3}^{0} [/tex]
Let m<M=b
[tex] \tan(b) = \frac{2}{3} [/tex]
[tex] \implies(b) = { \tan }^{ - 1} (\frac{2}{3})[/tex]
[tex]\implies(b) = {33.7}^{0} [/tex]
Hence
m<W=[tex]{56.3}^{0} [/tex]
m<X=[tex]{33.7}^{0} [/tex]
