Answer:
The area of the triangle is 18 square units.
Step-by-step explanation:
First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:
AB
BC
AC
Now we determine the area of the triangle by Heron's formula:
(1)
(2)
Where:
- Area of the triangle.
- Semiparameter.
If we know that , and , then the area of the triangle is:
The area of the triangle is 18 square units.
Hello again lol. I would think it would be two rooms don’t count me on this one but I think it is two
Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2() = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.
Well... to find 61 tens, just write 61 with the 1 in the tens place then all remaining places to the decimal get a zero (in this case that's just the ones place). So...
61 tens = 610
100 less than 610 is
610 - 100 = 510