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Sauron [17]
3 years ago
5

I have a right triangle. The hypotenuse is 12. The base is 7 and the other side is 9.7. The sin(x) is 0.014. Cos(x) is 0.99 and

tan(x) is 0.02. What is the measure of angle x?

Mathematics
1 answer:
Zanzabum3 years ago
6 0

Answer:

\sin(x)=\frac{\sqrt{95}}{12}=0.8122

\cos(x)=\frac{7}{12} \approx .5833  

\tan(x)=\frac{\sqrt{95}}{7} \approx 1.3924  

x=\sin^{-1}(\frac{\sqrt{95}}{12}) \approx 54.3147 degrees.

I don't know what you want to round to but I can tell you are rounding...

Step-by-step explanation:

Soh Cah Toa is the acronym we will use to help us solve this.

This says sine is opposite over hypotenuse.

It also says cosine is adjacent over hypotenuse.

And finally tangent is opposite over adjacent.

Note:  Opposite and adjacent can change depending on how you are looking at the triangle, like which angle you are referring to.

Let's look at x:

The side that is opposite, not touching, the angle whose measurement is x, is the side that has measurement \sqrt{95} \approx 9.7.

I see you already found this measurement. This is really good.  I might use the exact value for now and round at the end.

The side that is the hypotenuse no matter the angle you are referencing is always going to be opposite the angle whose measurement is 90 degrees. It is also the longest side (since it is opposite the largest angle).  This side has measurement 12.

The hypotenuse will be one of the sides touching your angle you are referencing.  The other side that is touching your angle is the adjacent side.  This side has measurement 7.

Anyways let's plug into the definitions we had above:

\sin(x)=\frac{\sqrt{95}}{12}  (O/H)

\cos(x)=\frac{7}{12}   (A/H)

\tan(x)=\frac{\sqrt{95}}{7}   (O/A)

Now we can solve anyone of these for x.

Take your pick.

\sin(x)=\frac{\sqrt{95}}{12}

x=\sin^{-1}(\frac{\sqrt{95}}{12})

x=54.3147 degrees.  

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If she has 280 dimes in piggy bank

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Answer:

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olga_2 [115]

<em><u>The solution is (4, 4)</u></em>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

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<em><u>Substitute eqn 2 in eqn 1</u></em>

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<em><u>Solve by quadratic equation</u></em>

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<em><u>Substitute x = 4 in eqn 2</u></em>

y = 2(4) - 4

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