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Radda [10]
3 years ago
7

On the packaging for a triangular sail, the edge measurements for the sail are listed as 7ft x 15ft x 17ft without unfurling the

sail, you want to determine if the sail forms a right triangle, an acute triangle or an obtuse triangle

Mathematics
2 answers:
kirza4 [7]3 years ago
5 0

The law of cosines can be helpful here. The largest angle (C) will be opposite the longest side (c). That law tells us ...

... c² = a² + b² -2ab·cos(C)

Then

... cos(C) = (a² +b² -c²)/(2ab)

We don't actually need the angle. We only need to know the sign of the cosine.

... cos(C) = (7² +15² -17²)/(2·7·15) = -15/210 . . . . negative sign, so C > 90°

The sail has the shape of an obtuse triangle.

_____

A triangle solver app on your calculator, phone, tablet, or computer can give you the result easily.

Furkat [3]3 years ago
3 0

The law of cosines can be helpful here. The largest angle (C) will be opposite the longest side (c). That law tells us ...


... c² = a² + b² -2ab·cos(C)


Then


... cos(C) = (a² +b² -c²)/(2ab)


We don't actually need the angle. We only need to know the sign of the cosine.


... cos(C) = (7² +15² -17²)/(2·7·15) = -15/210 . . . . negative sign, so C > 90°


The sail has the shape of an obtuse triangle.


_____


A triangle solver app on your calculator, phone, tablet, or computer can give you the result easily.




Read more on Brainly.com - brainly.com/question/11277503#readmore

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lidiya [134]

Answer:

41/35

Step-by-step explanation:

Simplify the following:

4/7 + 3/5

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5 0
3 years ago
A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th
Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner.  The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

\tan\theta=\frac{Perpendicular}{base}

Here, \theta=11.1^{\circ} ,Perpendicular AB = 4000

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BC=\frac{4000}{\tan 11.1^{\circ} }

BC=\frac{4000}{0.196} (approx)

BC=20408.16(approx)  

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

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3 years ago
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