Given that the sides of the acute triangle are as follows:
21 cm
x cm
2x cm
Stated that 21 cm is one of the shorter sides of the triangle2x is greater than x, so it follows that 2x MUST be the longest side
For acute triangles, the longest side must be less than the sum of the 2 shorter sides
Therefore, 2x < x + 21cm
2x – x < 21cm
x < 21cm
If x < 21cm, then 2x < 42cm
Therefore, the longest possible length for the longest side is 42cm
Answer:
Hello,
Step-by-step explanation:
![A=(1,2)\\B=(0,-1)\\\overrightarrow{AB}=((0,-1)-(1,2)=(-1,-3)\ ||\overrightarrow{AB}||^2=1+9=10\\\overrightarrow{BC}=((3,-2)-(0,-1)=(3,-1)\ ||\overrightarrow{BC}||^2=9+1=10\\\\Triangle\ is\ isosceles.\\\\\overrightarrow{AB}.\overrightarrow{BC}=(-1,-3)*\left[\begin{array}{c}3\\-1\end{array}\right] =-3+3=0\\\\Triangle \ is\ right.\\\\](https://tex.z-dn.net/?f=A%3D%281%2C2%29%5C%5CB%3D%280%2C-1%29%5C%5C%5Coverrightarrow%7BAB%7D%3D%28%280%2C-1%29-%281%2C2%29%3D%28-1%2C-3%29%5C%20%7C%7C%5Coverrightarrow%7BAB%7D%7C%7C%5E2%3D1%2B9%3D10%5C%5C%5Coverrightarrow%7BBC%7D%3D%28%283%2C-2%29-%280%2C-1%29%3D%283%2C-1%29%5C%20%7C%7C%5Coverrightarrow%7BBC%7D%7C%7C%5E2%3D9%2B1%3D10%5C%5C%5C%5CTriangle%5C%20is%5C%20isosceles.%5C%5C%5C%5C%5Coverrightarrow%7BAB%7D.%5Coverrightarrow%7BBC%7D%3D%28-1%2C-3%29%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%3D-3%2B3%3D0%5C%5C%5C%5CTriangle%20%5C%20is%5C%20right.%5C%5C%5C%5C)
Answer:
The actual length of 60 cm is modeled as 1 cm length.
Step-by-step explanation:
The diameter of the sweep of the main rotor of the helicopter is 18.6 meters i.e. 1860 cm.
Now, the toy model of the helicopter has a sweep of 31 cm.
Therefore, the scale in which the model is made is 1860 : 31 i.e. 60 : 1 ratio.
So, the actual length of 60 cm is modeled as 1 cm length. (Answer)
Answer:
- sin(2x) = -4/5
- cos(2x) = 3/5
- tan(2x) = -4/3
Step-by-step explanation:
It may be easiest to start with tan(2x).
tan(2x) = 2tan(x)/(1 -tan(x)²)
tan(2x) = 2(-1/2)/(1 -(-1/2)²) = -1/(3/4)
tan(2x) = -4/3 . . . . . still a 4th-quadrant angle
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Then cosine can be found from ...
cos(2x) = 1/√(tan(2x)² +1) = 1/√((-4/3)²+1) = √(9/25)
cos(2x) = 3/5
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Sine can be found from these two:
sin(2x) = cos(2x)tan(2x) = (3/5)(-4/3)
sin(2x) = -4/5
Since they are right triangles use Pythagorean Theorem.
a^2 + b^2 = c^2 where c is the hypotenuse.
a. 12^2 + b^2 = 13^2
144 + b^2 = 169
subtract 144 from both sides
b^2 = 25
take the square root of both sides
t = 5 This is also know as a Pythagorean Triple 5-12-13
b. a^2 + 9^2 = 12^2
a^2 + 81 = 144
subtract 81 from both sides
a^2 = 63
take the square root of both sides
a = √63
a = √(9 * 7)
a = 3√7
c. 6^2 + 9^2 = c^2
36 + 81 = c^2
117 = c^2
take the square root of each side
c = √117
c = √(9*13)
x = 3√13
