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2 years ago

15 POINTS!!!!

Mathematics

2 answers:

garik1379 [7]2 years ago

5 0

C because 20 over 5 equals 4, then 4+6 is 10

olganol [36]2 years ago

4 0

p/5 + 6 = 10

=> p/5 = 10-6

=> p/5 = 4

=> p = 5×4

=> p = 20

C. 20

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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

Ulleksa [173]

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°

Learn more about sine rule here:

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