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olga2289 [7]
3 years ago
15

How do you find the missing side??

Mathematics
2 answers:
dexar [7]3 years ago
6 0
The answer is 12 if I’m not mistaken
guapka [62]3 years ago
4 0

Answer:

\boxed{\ell = 12}.

Step-by-step explanation:

Let \ell denote the length of the missing side. Since \triangle ABC and \triangle AXY share one vertex and \overline{XY}\parallel \overline{BC}, these triangles are similar. This means that:

\dfrac{AX}{AB} = \dfrac{AY}{AC} \iff \dfrac{\ell}{18+\ell} = \dfrac{14}{14+21} \iff\dfrac{\ell}{18+\ell} = \dfrac{14}{35}.

We now multiply both sides by 18+\ell and solve the equation:

\dfrac{\ell}{18+\ell} = \dfrac{14}{35} \iff \ell = \dfrac{36}{5} + \dfrac{14}{35}\ell \iff \ell \left(1-\dfrac{14}{35}\right) = \dfrac{36}{5} \iff \dfrac{3}{5}\ell = \dfrac{36}{5}.

So we now simply get:

\boxed{\ell = \dfrac{36}{3} = 12}.

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Answer:

x=15

Step-by-step explanation:

<GCB=180-5x

<EBA=9x

<DAC=8x

The sum of the three exterior angles equals 360.

(180-5x)+9x+8x=360

180-5x+9x+8x=360

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x=15

6 0
3 years ago
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Please solve the following sum or difference identity.
xxTIMURxx [149]

Answer:

sin(A - B) = \frac{4}{5}

Step-by-step explanation:

Given:

sin(A) = \frac{24}{25}

sin(B) = -\frac{4}{5}

Need:

sin(A - B)

First, let's look at the identities:

sum: sin(A + B) = sinAcosB + cosAsinB

difference: sin(A - B) = sinAcosB - cosAsinB

The question asks to find sin(A - B); therefore, we need to use the difference identity.

Based on the given information (value and quadrant), we can draw reference triangles to find the simplified values of A and B.

sin(A) = \frac{24}{25}

cos(A) = \frac{7}{25}

sin(B) = -\frac{4}{5}

cos(B) = \frac{3}{5}

Plug these values into the difference identity formula.

sin(A - B) = sinAcosB - cosAsinB

sin(A - B) = (\frac{24}{25})(\frac{3}{5}) - (-\frac{4}{5})(\frac{7}{25})

Multiply.

sin(A - B) = (\frac{72}{125}) + (\frac{28}{125})

Add.

sin(A - B) = \frac{4}{5}

This is your answer.

Hope this helps!

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3 years ago
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