Answer:
The area of the triangle is [tex]\sqrt{3}[/tex]
Step-by-step explanation:
Given:
Coordinates D (0, 0), E (1, 1)
Angle ∠DEF = 60°
△DEF is a Right triangle
To Find:
The area of the triangle
Solution:
The area of the triangle is = [tex]\frac{1}{2}(base \times height)[/tex]
Here the base is Distance between D and E
calculation the distance using the distance formula, we get
DE = [tex]\sqrt{(0-1)^2 + (0-1)^2}[/tex]
DE =[tex]\sqrt{(-1) ^2 + (-1)^2[/tex]
DE = [tex]\sqrt{1+1}[/tex]
DE = [tex]\sqrt{2}[/tex]
Base = [tex]\sqrt{2}[/tex]
Height is DF
DF =[tex]tan(60^{\circ}) \times DE[/tex]
DF = [tex]\sqrt{3} \times DE[/tex]
DF = [tex]\sqrt{3} \times\sqrt{2}[/tex]
Now, the area of the triangle is
= [tex]\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})[/tex]
=[tex]\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})[/tex]
=[tex]\frac{1}{2}(2\sqrt{3} )[/tex]
=[tex]\sqrt{3}[/tex]
