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3 years ago

A skate ramp is 4.2 meters long. It rises up 1.7 meters. What’s is the angle of elevation to the nearest tenth of a degree?

1 answer:

4 0

Θ
=
arcsin
(
.7
4.2
)
≈
10
∘
Explanation:
We view the ramp as a right triangle. The hypotenuse is 4.2 and the vertical side .7, which is opposite the angle
θ
we seek.
sin
θ
=
.7
4.2
=
1
6
I'm going to finish the problem but I'll note if we were actually building the ramp we don't need to know the angle; this sine is sufficient.
θ
=
arcsin
(
1
6
)
θ
≈
10
∘
which I think is a pretty steep ramp for a wheelchair.
There will be another inverse sine that is the supplementary angle, around
170
∘
, but we can rule that out as a value for a ramp wedge angle.

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Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

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$\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9$

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

$125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9$

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