Answer:
a. tan∠FKG = √3
b. tan∠BKG = 1
c. [tex]tan\angle KDF = \dfrac{1}{\sqrt{3} }[/tex]
d. tan∠BCK = 1
e. The coordinates of the point g is (4, 4·√3)
Step-by-step explanation:
a. The trigonometric ratio for tan is given as follows;
[tex]tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length}[/tex]
Therefore, we have;
[tex]tan\angle FKG = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length} = tan(60^{\circ}) = \dfrac{sin(60^{\circ})}{cos(60^{\circ}) } = \dfrac{\dfrac{\sqrt{3} }{2} }{\dfrac{1}{2} } = \sqrt{3}[/tex]
b.
[tex]tan\angle BKG = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length} = tan(45^{\circ}) = \dfrac{sin(45^{\circ})}{cos(45^{\circ}) } = \dfrac{\dfrac{1 }{\sqrt{2} } }{\dfrac{1}{\sqrt{2} } } = 1[/tex]
c.
[tex]tan\angle KDF = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length} = tan(30^{\circ}) = \dfrac{sin(30^{\circ})}{cos(30^{\circ}) } = \dfrac{\dfrac{1 }{2 } }{\dfrac{\sqrt{3} }{2 } } = \dfrac{1}{\sqrt{3} }[/tex]
d. In right triangle, ΔBKC, ∠BCK = ∠BKC = 45°
∴ tan(∠BCK) = tan(45°) = 1
e. The x-coordinate of the point g is 4
The y-coordinate of the point g is 4 × tan(60°) = 4·√3
The coordinates of the point g is (4, 4·√3).
