Answer:
[tex]BD = 4[/tex]
[tex]BC =4\sqrt 3[/tex]
Step-by-step explanation:
Given
The attached triangle
Required
Find BD and BC
Solving BD
Considering angle at D, we have:
[tex]\cos(D) = \frac{Adjacent}{Hypotenuse}[/tex]
[tex]\cos(60) = \frac{BD}{8}[/tex]
Solve for BD
[tex]BD = 8 * \cos(60)[/tex]
[tex]\cos(60) = 0.5[/tex] So:
[tex]BD = 8 * 0.5[/tex]
[tex]BD = 4[/tex]
To solve for BC, we make use of Pythagoras theorem
[tex]CD^2 = BC^2 + BD^2[/tex]
This gives
[tex]8^2 = BC^2 + 4^2[/tex]
[tex]64 = BC^2 + 16[/tex]
Collect like terms
[tex]BC^2 =64-16[/tex]
[tex]BC^2 =48[/tex]
Take square roots
[tex]BC =\sqrt{48[/tex]
Expand
[tex]BC =\sqrt{16*3[/tex]
Split
[tex]BC =\sqrt{16}*\sqrt 3[/tex]
[tex]BC =4\sqrt 3[/tex]
