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Arlecino [84]
3 years ago
6

Please show all work

Mathematics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

<h2>I. 336 ft²</h2>

Step-by-step explanation:

We have

two right triangle with legs 6ft and 8ft

one rectangle 8ft × 12ft

one rectangle 6ft × 12ft

one rectangle 10ft × 12ft

The formula of an area of a right triangle:

A=\dfrac{ab}{2}

a,b - legs

Substitute:

A_1=\dfrac{(6)(8)}{2}=(3)(8)=24\ ft^2

The formula of an area of a rectangle:

A=lw

l,w - length, width (dimensions of rectangle)

Substitute:

A_2=(6)(12)=72\ ft^2\\\\A_3=(8)(12)=96\ ft^2\\\\A_4=(10)(12)=120\ ft^2

The Surface Area:

S.A.=2A_1+A_2+A_3+A_4\\\\S.A.=(2)(24)+72+96+120=336\ ft^2

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The plane will end up flying 5.02°.

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Solution:

Use the cosine formula,

a^{2}=b^{2}+c^{2}-2 b c \cos A

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\mathrm{R}^{2}=700^{2}+80^{2}-(2 \times 700 \times 80 \times \cos 45^\circ)

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This is the grouped speed of the aircraft.

To find θ use sine rule.

$\frac{\sin C}{c}=\frac{\sin A}{a}

$\frac{\sin \theta}{80}=\frac{\sin 45}{645.91}

Do cross multiplication, we get

${\sin \theta}}=\frac{\sin 45}{645.91}\times 80

${\sin \theta}}=\frac{\frac{1}{\sqrt{2} } }{645.91}\times 80

sin θ = 0.0875

θ = 5.02°

This is known as the drift angle and is the correction the pilot should apply to remain on course.  

The heading is the direction the aircraft's nose is pointing which is

The track is the actual direction over the ground which is θ = 5.02°

An alternative method to this would be to separate each vector into vertical and horizontal components and add.

The resultant can be found using Pythagoras.

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