Law of cosines
:
The law of cosines establishes:

general guidelines:
The law of cosines is used to find the missing parts of an oblique triangle (not rectangle) when either the two-sided measurements and the included angle measure are known (SAS) or the lengths of the three sides (SSS) are known.
Law of the sines:
In ΔABC is an oblique triangle with sides a, b, and c, then:

The law of the sines is the relation between the sides and angles of triangles not rectangles (obliques). It simply states that the ratio of the length of one side of a triangle to the sine of the angle opposite to that side is equal for all sides and angles in a given triangle.
General guidelines:
To use the law of the sines you need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an opposite angle of one of them (SSA).
The ambiguous case
:
If two sides and an angle opposite one of them is given, three possibilities may occur.
(1) The triangle does not exist.
(2) Two different triangles exist.
(3) Exactly a triangle exists.
If we are given two sides and an included angle of a triangle or if we are given 3 sides of a triangle, we can not use the law of the sines because we can not establish any proportion where sufficient information is known. In these two cases we must use the law of cosines
Answer:
27 degrees
Step-by-step explanation:
We are given that Angle AEF is 63. We can also see that EAF is a right angle. Since angles in a triangle add up to 180, we can use this to solve for AFE:
Angle AEF + EAF + AFE = 180
63 + 90 + AFE = 180
153 + AFE = 180
AFE = 27
<em>Or we can solve it another way</em>
Since EAF is a right angle, the other angles are complementary (they add up to 90) so...
Angle AEF + AFE = 90
63 + AFE = 90
AFE = 27
Number 7 is C, im not sure about 6 though
Answer:
The data table is attached below.
Step-by-step explanation:
The average of a set of data is the value that is a representative of the entire data set.
The formula to compute averages is:

Compute the average for drop 1 as follows:
![\bar x_{1}=\frac{1}{3}\times[10+11+9]=10](https://tex.z-dn.net/?f=%5Cbar%20x_%7B1%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%5B10%2B11%2B9%5D%3D10)
Compute the average for drop 2 as follows:
![\bar x_{2}=\frac{1}{3}\times[29+31+30]=30](https://tex.z-dn.net/?f=%5Cbar%20x_%7B2%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%5B29%2B31%2B30%5D%3D30)
Compute the average for drop 3 as follows:
![\bar x_{3}=\frac{1}{3}\times[59+58+61]=59.33](https://tex.z-dn.net/?f=%5Cbar%20x_%7B3%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%5B59%2B58%2B61%5D%3D59.33)
Compute the average for drop 4 as follows:
![\bar x_{4}=\frac{1}{3}\times[102+100+98]=100](https://tex.z-dn.net/?f=%5Cbar%20x_%7B4%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%5B102%2B100%2B98%5D%3D100)
Compute the average for drop 5 as follows:
![\bar x_{5}=\frac{1}{3}\times[122+125+127]=124.67](https://tex.z-dn.net/?f=%5Cbar%20x_%7B5%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%5B122%2B125%2B127%5D%3D124.67)
The data table is attached below.
It's A. If U=16, 6+10=16
and if you plug in the numbers for the second equation it's 36=2(10)+16