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3 years ago

I need help on this

Mathematics

1 answer:

Slav-nsk [51]3 years ago

6 0

Based in the given figure, we are being asked to solve the area and the perimeter of the semicircle. As we evaluate the problem, we can get a radius measurement from the rectangle. Hence, the radius is half of 4 inches which is 2 inches and since we know the formula for solving the area of a circle which is Area = pi*r², dividing the result by two, we able to get the area of the half of a circle which is equivalent to the area of the semicircle.
Area of circle = pi*r²
Area of circle = 3.14*(2)² = 12.56
Therefore, area of semicircle is = (1/2) 12.56
Area of semicircle = 6.28 inches²

Solving for the perimeter:
Semicircle Perimeter = 1/2 * pi* d+d where d is the diameter (diameter = 4 inches)
Semicircle Perimeter = 1/2 * 3.14*4+4
Semicircle Perimeter = 10.28 inches

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David and Lacey are 2000 feet apart on a coastline. They each look at the same boat in the water. The angle between the coastlin

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By using the given distances and directions, we have;

Part A: 1083.5 feet

Part B: 1921.37 feet

Part C: 58.05°

<h3>How can the distances and directions be calculated?</h3>

The distance between David and Lacey = 2000 feet

David's angle to the boat = 70°

Lacey's angle to the boat = 32°

Part A

The distance of the boat from David is found as follows;

Imaginary lines drawn from the boat to David and then to Lacey form a triangle.

In the triangle, let A = 70°

B = 32°

Therefore by the sum of angles in a triangle, we have;

C = 180° - (70° + 32°) = 78°

By using sine rule we have;

$\frac{a}{sin(A)} = \frac{b}{sin( B )} = \frac{c}{sin(C)}$

David's distance from the boat, b, is therefore;

$\frac{b}{sin( 32 )} = \frac{2000}{sin(78)}$

The angle subtended by the coastline, C, is therefore;

$b = \frac{2000}{sin(78)} \times sin( 32 ) = 1083.5$

David's distance from the boat is 1083.5 feet

Part B

The distance between the boat and Lacey is found as follows;

$\mathbf{ \frac{a}{sin( 70)} }= \frac{2000}{sin(78)}$

$a = \frac{2000}{sin(78)} \times sin( 70 ) = 1921.37$

Lacey's distance from the boat is 1921.37 feet

Part C

When b = 1200 feet, we have;

Finding the vertical distance of the boat from the coastline, we have;

1083.5 × sin(70) = 1018.17

We have;

$\frac{1200}{sin(90)} = \frac{1018.17}{sin( B' )}$

${sin( B' )} = \frac{sin(90)}{1200} \times 1018.17$

The angle between the coastline and his view to the boat, B', is therefore;

B' = arcsine (1018.17÷1200) = 58.05°

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3 years ago

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