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3 years ago

How do you solve this using special right triangle this is geometry

Mathematics

1 answer:

svlad2 [7]3 years ago

3 0

Hello from MrBillDoesMath!

Answer:

11.1

Discussion:

cos(42) = x / 15 => multiply both sides by 15

x = 15 * cos(42) => as cos(42) = .743 approx

x = 15 * (.743)

x = 11.15

Thank you,

MrB

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The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

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Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

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Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

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The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

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To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

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Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

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$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

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