Geometry help me please

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

The first blank is segment addition postulate and the second blank is substitution property of equality

You might be interested in

<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"

qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that $\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)$

Also, $-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)$

This means our original expression is equal to:

$\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]$

Using the cosine addition formula, which states $\cos(a+b)=\cos a \cos b-\sin a \sin b$ , we get this itself is equal to:

$\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)$

Since $\sin^{2} \theta+\cos^{2} \theta=1$ , we know that:

$\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}$

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

$\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}$