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3 years ago

Can anyone please help me with this geometry question?

Mathematics

1 answer:

Ainat [17]3 years ago

8 0

Answer:

Step-by-step explanation:

let base be x and height be 3x

we know area of parallelogram is

base*height

ie x*3x=675

3x^2=675

x^2=675/3

x^2=225

x= square root of 225

x=15

therefore base =15

height = 45

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There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of

Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)

Where:

$AC$ - The distance from A to C, measured in miles.

$AB$ - The distance from A to B, measured in miles.

$BC$ - The distance from B to C, measured in miles.

$\theta$ - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

$AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta$

$\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}$

$\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right)$ (2)

If we know that $BC = 7\,mi$ , $AB = 8\,mi$ , $AC = 5\,mi$ , then the bearing from island B to island C:

$\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]$

$\theta \approx 38.213^{\circ}$

The bearing needed to navigate from island B to island C is approximately 38.213º.

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