The figure is a parallelogram.

Mathematics

2 answers:

spin [16.1K]3 years ago

7 0

The answer will be C, 40 degrees

Nonamiya [84]3 years ago

4 0

I hope this helps you

m (ACD)=m (ABD)

4x+4=6x-14

2x=18

x=9

m (ACD)=4x+4 =4.9+4=40

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Answer:

See the step by step solution

Step-by-step explanation:

Consider a more general case of the form $a\sin(x)+b\cos(x)$ . We want to express it as a sine term only by using the following formula $\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

The trick consists on find a number $\beta \in [0,2\pi]$ such that $\cos(\beta) = a, \sin(\beta)=b$ . But, note that since $\cos^2(\beta)+\sen^2(\beta)=1$ it is most likely that $a^2+b^2neq 1$ . Then, we will use the following equations:

$\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B$

Then, problem turns out to be

$a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)$ ,

where $\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})$

b) The amplitude is $\sqrt[]{15^2+4^2} = \sqrt[]{241}$ . Since $2\pi rad/s = 1 Hz$ the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.