Answer:
The area will be:
[tex]A=18(1+\sqrt{3}) \: u^{2}[/tex] or [tex]A=49.18 \: u^{2}[/tex].
Step-by-step explanation:
Using the definition of tangent, we have:
[tex]tan(\alpha)=\frac{6}{\vec{AB}}[/tex]
We know that tan(α) = 1, then we can find AB
[tex]1=\frac{6}{\vec{AB}}[/tex]
[tex]\bar{AB}=6\: u[/tex]
u means any unit.
Now, we need to find the distance BC. If the angle ∠D is 60°, then ∠C must be 30°. Using the tangent definition in the triangle BCD we have:
[tex]tan(30)=\frac{6}{\bar{BC}}[/tex]
[tex]\bar{BC}=\frac{6}{tan(30)}[/tex]
[tex]\bar{BC}=\frac{6}{1/\sqrt{3}}[/tex]
[tex]\bar{BC}=6\sqrt{3} \: u[/tex]
So, the base of the triangle will be:
[tex]b=\bar{AB}+\bar{BC}=6+6\sqrt{3}=6(1+\sqrt{3}) \: u[/tex]
The area of a triangle is given by the following equation:
[tex]A=\frac{b*h}{2}[/tex]
- b is the base ([tex]b=6(1+\sqrt{3})\: u[/tex])
- h is the height (h=6 u)
[tex]A=\frac{6(1+\sqrt{3})*6}{2}[/tex]
[tex]A=18(1+\sqrt{3})[/tex]
Therefore, the area will be [tex]A=49.18 \: u^{2}[/tex].
I hope it helps you!
