It would be 32.15 meters
You can draw a right triangle (Observe the figure attached. It is not drawn to scale), where "x" is the the amount of meters the street rises over a horizontal distance of 120 meters.
You need to use the following Trigonometric Identity:
In this case, you can identify that:
Then, knowing these values, you can substitute them into:
And finally, you must solve for "x" in order to find its value.
Those wind chill formulas are REALLY complex.
Attached are the formulas for wind chill (F and C) plus the OLD wind chill formula.
Here's a wind chill calculator I wrote. http://www.1728.org/windchil.htm
(I figured you might need all the help you can get.)
Answer:
35.09°
Step-by-step explanation:
You can use cos theta to find the value of x.
Let us solve now.
They have already given the lengths of the hypotenuse and adjacent.
Hypotenuse = 33
Adjacent = 27
Let us solve now.
Cos x = Adjacent ÷ Hypotenuse
Cos x = 27 ÷ 33
Cos x = 0.8181
x = Cos⁻¹ 0.8181
x = 35.09°
Hope this helps you :-)
Let me know if you have any other questions :)
<h3>
Answer: C) 142 degrees</h3>
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Explanation:
Extend segment MN such that it intersects side ST. Mark the intersection as point A. See the diagram below.
We're given that angle MNT is 72 degrees. The angle TNA is equal to 180-(angle MNT) = 180 - 72 = 108 degrees, since angles MNT and TNA add to 180.
For now, focus entirely on triangle TNA. We see from the diagram that T = 34 and we just found that N = 108. Let's find angle A
A+N+T = 180
A+108+34 = 180
A+142 = 180
A = 180-142
A = 38
So angle NAT is 38 degrees.
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Since segment MA is an extension of MN, and because MN || SQ, this means MA is also parallel to SQ.
We found at the conclusion of the last section that angle NAT was 38 degrees. Angles QST and NAT are corresponding angles. They are congruent since MA || SQ. This makes angle QST to also be 38 degrees
----------------------------
The angles QSR and QST are a linear pair, so they are supplementary
(angle QSR) + (angle QST) = 180
angle QSR = 180 - (angle QST)
angle QSR = 180 - 38
angle QSR = 142 degrees