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3 years ago

In ΔTUV, the measure of ∠V=90°, the measure of ∠T=16°, and VT = 4.9 feet. Find the length of TU to the nearest tenth of a foot.

Mathematics

1 answer:

RSB [31]3 years ago

6 0

<u>Given</u>:

Given that TUV is a right triangle.

The measure of ∠T is 16°

The length of VT is 4.9 feet.

We need to determine the length of TU.

<u>Length of TU:</u>

The length of TU can be determined using the trigonometric ratio.

Thus, we have;

$cos \ \theta=\frac{adj}{hyp}$

where $\theta=16^{\circ}$ , adj = TV and hyp = TU

Thus, we get;

$cos \ 16^{\circ}=\frac{4.9}{TU}$

Simplifying, we get;

$TU=\frac{4.9}{cos \ 16^{\circ}}$

$TU=\frac{4.9}{0.9613}$

$TU=5.097$

Rounding off to the nearest tenth, we get;

$TU=5.1$

Thus, the length of TU is 5.1 feet.

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A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

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mamaluj [8]

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Triangle ABC is translated according to the rule (x,y) arrow (x+2,y-8) if the coordinates of the pre image of point B are (4,-5)

kherson [118]

Answer: the coordinates of B' are (6, -13)

Justification:

1) Translating triangle ABC according to the rule (x,y) → (x + 2, y - 8) means that every single point of the triangle will be translated two units to the righ (x + 2) and 8 units downward (y - 8).

2) To find the coordinates of anay image you have to add up 2 to the x coordinate (x + 2) and subtract 8 from the y-coordinate (y - 8).

3) Peforming those simple operations to the coordinates of the point B (4, -5), you will obtain the point B':

* x' = x + 2 = 4 + 2 = 6
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Answer: the coordinates of B' are (6, -13)

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