Problem 1
x = measure of angle N
2x = measure of angle M, twice as large as N
3(2x) = 6x = measure of angle O, three times as large as M
The three angles add to 180 which is true of any triangle.
M+N+O = 180
x+2x+6x = 180
9x = 180
x = 180/9
x = 20 is the measure of angle N
Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.
<h3>Answers:</h3>
- Angle M = 40 degrees
- Angle N = 20 degrees
- Angle O = 120 degrees
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Problem 2
n = number of sides
S = sum of the interior angles of a polygon with n sides
S = 180(n-2)
2700 = 180(n-2)
n-2 = 2700/180
n-2 = 15
n = 15+2
n = 17
<h3>Answer: 17 sides</h3>
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Problem 3
x = smaller acute angle
3x = larger acute angle, three times as large
For any right triangle, the two acute angles always add to 90.
x+3x = 90
4x = 90
x = 90/4
x = 22.5
This leads to 3x = 3*22.5 = 67.5
<h3>Answers:</h3>
- Smaller acute angle = 22.5 degrees
- Larger acute angle = 67.5 degrees
Answer:
B or $12.50
Step-by-step explanation:
To get the answer, you have to solve x from the equation 4.5x = 56.25
To solve, you have to divide 4.5 from each side, making the equation look like,
x = 12.5
Hope this helps!!
Plz let me know if I'm wrong...
Painting a mural
Doing charity work like playing a concert
Design
you will need to use the law of cosines since this picture does not indicate that this is a right triangle
c^2 = a^2 + b^2 – 2ab cos C,
16^2 = 17^2+8^2 -2*17*8*cosC
256=289-272cosC
-33=-272cosC
33/272 = cosC
cos^-1 (33/272)=C taking the inverse cos
C=83.0 (to the nearest tenth)
b^2 = a^2 + c^2 – 2ac cos B,
8^2 = 17^2 +16^2 -2*17*16cosB
64=289-544cosB
-225=-544cosB
225/544=cosB
cos^(-1) =B
B=65.5
A=180-83-65.6
A=31.4
