xand y are light house. y being 20km due east of x and from a ship due south of x the bearing of y was 055°. what is the distanc
e of the ship from y and the distance of the ship from x?
1 answer:
Answer:
I) |xz| ≈ 28.6 km
II) |yz| ≈ 34.8 km
Step-by-step explanation:
Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)
|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°
Using Trigonometric ratio - SOHCAHTOA
I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20
|xz| = 20 * Tan 55 = 20 * 1.428
|xz| = 28.56 km
|xz| ≈ <u>28.6 km</u>
<u />
II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|
|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°
|yz| = 20 ÷ 0.574 = 34.84 km
|yz| ≈ <u>34.8 km</u>
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Answer:
shaded area = 113.64 cm²
Step-by-step explanation:
To find the shaded area, subtract the area of the circle from the area of the triangle.
Area of a triangle = 1/2 x base x height
⇒ area of triangle = 1/2 x 25 x 21.4 = 267.5 cm²
Area of a circle = (where r is the radius)
From inspection, we can see that the diameter of the circle = 14 cm
Therefore, as the diameter = 2r, then r = 14 ÷ 2 = 7 cm
⇒ area of circle = 3.14 x 7² = 153.86 cm²
Shaded area = area of triangle - area of circle
⇒ shaded area = 267.5 - 153.86 = 113.64 cm²