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nevsk [136]
3 years ago
6

A ship sails 95 km on a bearing of 140 degrees,than further on 102km on a bearing of 260degrees,and than returns directly to its

starting point.Find the length and bearing of the return journey

Mathematics
1 answer:
sammy [17]3 years ago
7 0

Answer:

1. 98.7 km

2. 303.5 degree

Step-by-step explanation:

Using the alternate angle, the angle at B will be 50 + 10 = 60 degree.

To calculate the length AC of the returning journey, use cosine formula to calculate it.

To find the bearing of the returning journey, use sine rule to calculate it.

Please find the attached file for the solution

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Answer:

<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>

Step-by-step explanation:

Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units

From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.

By the definition of rhombus, diagonals meet at right angles.

Implies that PQ = QA

x+2 = 3x - 14

x-3x=-14-2

-2x=-16

2x = 16

dividing by 2 on both sides, we will get,

x =\frac{16}{2}

<h3>∴ x=8</h3><h3>Since Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles we can equate x+2 = 3x-14 to find the value of x.</h3>

The line segment \overrightarrow{PA}=\overrightarrow{PQ}+\overrightarrow{QA}

\overrightarrow{PA}=x+2+3x-14

=4x-12

=4(8)-12 ( since x=8)

=32-12

=20

<h3>∴ \overrightarrow{PA}=20 units</h3>
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see image

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