Answer:
Part A) ![sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}](https://tex.z-dn.net/?f=sin%28%5Calpha%29%3D%5Cfrac%7B4%7D%7B7%7D%2C%5C%20cos%28%5Cbeta%29%3D%5Cfrac%7B4%7D%7B7%7D)
Part B) ![tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}](https://tex.z-dn.net/?f=tan%28%5Calpha%29%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B33%7D%7D%2C%5C%20tan%28%5Cbeta%29%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B33%7D%7D)
Part C) ![sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}](https://tex.z-dn.net/?f=sec%28%5Calpha%29%3D%5Cfrac%7B7%7D%7B%5Csqrt%7B33%7D%7D%2C%5C%20csc%28%5Cbeta%29%3D%5Cfrac%7B7%7D%7B%5Csqrt%7B33%7D%7D)
Step-by-step explanation:
Part A) Find ![sin(\alpha)\ and\ cos(\beta)](https://tex.z-dn.net/?f=sin%28%5Calpha%29%5C%20and%5C%20cos%28%5Cbeta%29)
we know that
If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle
In this problem
---> by complementary angles
so
![sin(\alpha)=cos(\beta)](https://tex.z-dn.net/?f=sin%28%5Calpha%29%3Dcos%28%5Cbeta%29)
Find the value of
in the right triangle of the figure
---> opposite side divided by the hypotenuse
simplify
therefore
Part B) Find ![tan(\alpha)\ and\ cot(\beta)](https://tex.z-dn.net/?f=tan%28%5Calpha%29%5C%20and%5C%20cot%28%5Cbeta%29)
we know that
If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle
In this problem
---> by complementary angles
so
![tan(\alpha)=cot(\beta)](https://tex.z-dn.net/?f=tan%28%5Calpha%29%3Dcot%28%5Cbeta%29)
<em>Find the value of the length side adjacent to the angle alpha</em>
Applying the Pythagorean Theorem
Let
x ----> length side adjacent to angle alpha
![14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132](https://tex.z-dn.net/?f=14%5E2%3Dx%5E2%2B8%5E2%5C%5Cx%5E2%3D14%5E2-8%5E2%5C%5Cx%5E2%3D132)
![x=\sqrt{132}\ units](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B132%7D%5C%20units)
simplify
![x=2\sqrt{33}\ units](https://tex.z-dn.net/?f=x%3D2%5Csqrt%7B33%7D%5C%20units)
Find the value of
in the right triangle of the figure
---> opposite side divided by the adjacent side angle alpha
simplify
![tan(\alpha)=\frac{4}{\sqrt{33}}](https://tex.z-dn.net/?f=tan%28%5Calpha%29%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B33%7D%7D)
therefore
![tan(\alpha)=\frac{4}{\sqrt{33}}](https://tex.z-dn.net/?f=tan%28%5Calpha%29%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B33%7D%7D)
![tan(\beta)=\frac{4}{\sqrt{33}}](https://tex.z-dn.net/?f=tan%28%5Cbeta%29%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B33%7D%7D)
Part C) Find ![sec(\alpha)\ and\ csc(\beta)](https://tex.z-dn.net/?f=sec%28%5Calpha%29%5C%20and%5C%20csc%28%5Cbeta%29)
we know that
If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle
In this problem
---> by complementary angles
so
![sec(\alpha)=csc(\beta)](https://tex.z-dn.net/?f=sec%28%5Calpha%29%3Dcsc%28%5Cbeta%29)
Find the value of
in the right triangle of the figure
Find the value of ![cos(\alpha)](https://tex.z-dn.net/?f=cos%28%5Calpha%29)
---> adjacent side divided by the hypotenuse
simplify
![cos(\alpha)=\frac{\sqrt{33}}{7}](https://tex.z-dn.net/?f=cos%28%5Calpha%29%3D%5Cfrac%7B%5Csqrt%7B33%7D%7D%7B7%7D)
therefore