Right △EFG has its right angle at F, EG=8, and FG=6.

What is the value of the trigonometric ratio of an angle of the triangle?

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

40 is the final : )