Answer:
[tex]VW = 6[/tex]
[tex]WX = 2\sqrt{3}[/tex]
[tex]\angle V = 30[/tex]
Step-by-step explanation:
Given
[tex]VX = 4\sqrt{3[/tex]
[tex]\angle X = 60^\circ[/tex]
See attachment for right triangle
Required
Solve for VW, WX and [tex]\angle V[/tex]
First, calculate VW
To do this, we make use of:
[tex]sin(\theta) = \frac{Opposite}{Hypotenuse}[/tex]
In this case,
[tex]sin(60) = \frac{VW}{4\sqrt{3}}[/tex]
Make VW the subject
[tex]VW = 4\sqrt{3} * sin(60)[/tex]
In radical form:
[tex]sin(60) = \frac{\sqrt{3}}{2}[/tex]
[tex]VW = 4\sqrt{3} * \frac{\sqrt{3}}{2}[/tex]
[tex]VW = \frac{4\sqrt{3} * \sqrt{3}}{2}[/tex]
[tex]VW = \frac{4*3}{2}[/tex]
[tex]VW = \frac{12}{2}[/tex]
[tex]VW = 6[/tex]
To solve for WX, we make use of Pythagoras Theorem, we make use of:
[tex]VX^2 = VW^2 + WX^2[/tex]
[tex](4\sqrt{3})^2 = 6^2 + WX^2[/tex]
[tex]16*3 = 36 + WX^2[/tex]
[tex]48= 36 + WX^2[/tex]
Subtract 36 from both sides
[tex]48 - 36 = 36 - 36 + WX^2[/tex]
[tex]48 - 36 = WX^2[/tex]
[tex]12= WX^2[/tex]
Take the square root of both sides
[tex]\sqrt{12}= WX[/tex]
[tex]\sqrt{4*3}= WX[/tex]
[tex]\sqrt{4}*\sqrt{3}= WX[/tex]
[tex]2\sqrt{3}= WX[/tex]
[tex]WX = 2\sqrt{3}[/tex]
Solving [tex]\angle V[/tex]
To do this, we make use of:
[tex]\angle V + \angle X + 90 = 180[/tex]
Make V the subject
[tex]\angle V = 180 - 90 - \angle X[/tex]
[tex]\angle V = 180 - 90 - 60[/tex]
[tex]\angle V = 30[/tex]
