Answer:
cos∠FHE = 4/5
Explanation:
* Lets revise the trigonometry functions
- In ΔABC
# m∠B = 90°
# Length of AB = a , length of BC = b and length of AC = c
# The trigonometry functions of angle C are
- sin(C) = a/c ⇒ opposite side to ∠C ÷ the hypotenuse
- cos(C) = b/c ⇒ adjacent side to ∠C ÷ the hypotenuse
- tan(c) = a/b ⇒ opposite side to ∠C ÷ adjacent side to ∠C
* Now lets solve the problem
- To find cos∠FHE , we must find the length of the adjacent
side HF and the length of the hypotenuse HE
- We can find the length of HF from ΔFGH
- In ΔFGH
∵ m∠G = 90°
∵ GF = √8
∵ m∠GHF = 45°
∵ sin∠GHF = GF/HF
∴ sin45° = √8/HF ⇒ by using cross-multiplication
∴ HF × sin45° = √8
∵ sin45° = 1/√2
∴ HF × 1/√2 = √8 ⇒ multiply each side by √2
∴ HF = √2 × √8 = √16 = 4
* Lets find the length of HE from ΔHFE
- In ΔHFE
∵ m∠HFE = 90°
∵ EF = 3 ⇒ given
∵ HF = 4
- By using Pythagoras theorem
∵ HE = √(FH² + FE²)
∴ HE = √(4² + 3²) = √(16 + 9) = √25 = 5
* Now we can find cos∠FHE
∵ cos∠FHE = HF/HE
∵ HF = 4 and HE = 5
∴ cos∠FHE = 4/5