Answer:
I. sec G = 4/3
II. cot G = (3√7)/7
III. sin E = ¾
Step-by-step explanation:
Given
Right angled triangle
EG = 8
FG = 6
Required
sec G
cot G
sin E
First, the length of the three sides of the triangle must be completed.
From the attachment below, we can deduce that (by using Pythagoras theorem)
Hyp² = Adj² + Opp²
EG² = FG² + EF²
Substitute EG = 8 and FG = 6
8² = 6² + EF²
64 = 36 + EF²
Collect like terms
EF² = 64 - 36
EF² = 28
Take square root of both sides
√EF² = √28
EF = √28
EF = √(4 * 7)
EF = √4 * √7
EF = 2 * √7
EF = 2√7.
Now, the trigonometric functions can be solved.
i. sec G.
sec is the reciprocal of cos .
So, we need to solve first for cos G
cos G = Adj/Hyp
Here, adj = 6 and hyp = 8
So,
cos G = 6/8
cos G = ¾
So,
sec G = 1/(¾)
sec G = 4/3
ii. cot G
cot is the reciprocal of tan
So, we need to solve first for tan G
tan G = Opp/Adj
Here, adj = 6 and opp = 2√7
So,
tan G = 2√7/6
tan G = √7/3
So,
cot G = 1/(√7/3)
cot G = 3/√7
Rationalise
cot G = 3/√7 * √7/√7
cot G = (3√7)/7
iii. sin E
sin E = Opp/Hyp
Here, Opp = 6
Hyp = 8
So,
sin E = 6/8
sin E = ¾
