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earnstyle [38]
3 years ago
6

Please help... ABCD is a quadrilateral​

Mathematics
1 answer:
IgorC [24]3 years ago
3 0
A ———————————————————
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Given a similar pair of triangles, find the missing length x and the missing angles A, B, and C.
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Answer:

x=81.25, A=65, B=94, C=21

Step-by-step explanation:

corresponding angles of similar triangles are equal, hence angle measurements of A, B, and C. To solve for x, set up a ratio (60/48=x/65) and solve

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The angle 0 lies in Quadrant II . <br>Cos 0 = <br><img src="https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%7D%7B3%7D%20" id="TexF
mel-nik [20]

Answer:

\tan(\theta)=\frac{-\sqrt{5}}{2}

Step-by-step explanation:

Since we are in quadrant 2, sine is positive.  Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely \sin^2(\theta)+\cos^2(\theta)=1.

If given \cos(\theta)=\frac{-2}{3}, then we have \sin^2(\theta)+(\frac{-2}{3})^2=1 by substitution of \cos(\theta)=\frac{-2}{3}.

Let's solve:

\sin^2(\theta)+(\frac{-2}{3})^2=1 for \sin(\theta).

\sin^2(\theta)+\frac{4}{9}=1

Subtract 4/9 on both sides:

\sin^2(\theta)=1-\frac{4}{9}

Simplify:

\sin^2(\theta)=\frac{5}{9}

Square root both sides:

\sin(\theta)=\sqrt{\frac{5}{9}}

\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}

\sin(\theta)=\frac{\sqrt{5}}{3}

===========

\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}

Multiplying top and bottom by 3 gives:

\tan(\theta)=\frac{\sqrt{5}}{-2}

I'm going to move the factor of -1 to the top:

\tan(\theta)=\frac{-\sqrt{5}}{2}

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It should be -3.125%

Step-by-step explanation:

(124 - 128)/128 × 100

-4/128 × 100 = -3.125%

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use calculator

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