Answer:
17) Ф = 57.99 ≅ 58°
18) Ф = 20.10 ≅ 21°
Step-by-step explanation:
* Lets revise the trigonometry functions
- In any right angle triangle:
# The side opposite to the right angle is called the hypotenuse
# The other two sides are called the legs of the right angle
* If the name of the triangle is ABC, where B is the right angle
∴ The hypotenuse is AC
∴ AB and BC are the legs of the right angle
- ∠A and ∠C are two acute angles
- For angle A
# sin(A) = opposite/hypotenuse
∵ The opposite to ∠A is BC
∵ The hypotenuse is AC
∴ sin(A) = BC/AC
# cos(A) = adjacent/hypotenuse
∵ The adjacent to ∠A is AB
∵ The hypotenuse is AC
∴ cos(A) = AB/AC
# tan(A) = opposite/adjacent
∵ The opposite to ∠A is BC
∵ The adjacent to ∠A is AB
∴ tan(A) = BC/AB
* Now lets solve the problems
17) In Δ ACB
∵ m∠C = 90°
∵ BC = 16 units ⇒ opposite to angle Ф
∵ AC = 10 units ⇒ adjacent to angle Ф
∵ tanФ = opposite/adjacent
∴ tanФ = BC/AC
∴ tanФ = 16/10 = 8/5
- To find angle Ф find the inverse of tan (tan^-1)
∴ Ф = tan^-1 8/5 = 57.99 ≅ 58°
18) In Δ ACB
∵ m∠C = 90°
∵ BC = 5.6 units ⇒ opposite to angle Ф
∵ AC = 15.3 units ⇒ adjacent to angle Ф
∵ tanФ = opposite/adjacent
∴ tanФ = BC/AC
∴ tanФ = 5.6/15.3 = 56/153
- To find angle Ф find the inverse of tan (tan^-1)
∴ Ф = tan^-1 56/153 = 20.10 ≅ 21°
