Question

Question 10 part a and b please I need help due tmmr

Answer 1

∠ E ≈ 38.9° , ∠ G ≈ 79.6°

Explanation:

given 3 sides of a triangle, to find any of the angles use the Cosine rule

(A)

cos E =
`(f^2+g^2-e^2)/(2fg)`

where f is the fide opposite ∠ F , g is the side opposite ∠ G and

e is the side opposite ∠ E

here f = EG = 42 , g = EF = 47 and e = FG = 30 , then

cos E =
`(42^2+47^2-30^2)/(2(42)(47))` =
`(1764+2209-900)/(3948)` =
`(3073)/(3948)` , then

∠ E =
`cos^(-1)` (
`(3073)/(3948)` ) ≈ 38.9° ( to the nearest tenth )

(b)

similarly, using e, f , g as above in part A

cos G =
`(e^2+f^2-g^2)/(2ef)` =
`(30^2+42^2-47^2)/(2(30)(42))` =
`(900+1764-2209)/(2520)` =
`(455)/(2520)` , then

∠ G =
`cos^(-1)` (
`(455)/(2520)` ) ≈ 79.6° ( to the nearest tenth )