So lets get to the problem
<span>165°= 135° +30° </span>
<span>To make it easier I'm going to write the same thing like this </span>
<span>165°= 90° + 45°+30° </span>
<span>Sin165° </span>
<span>= Sin ( 90° + 45°+30° ) </span>
<span>= Cos( 45°+30° )..... (∵ Sin(90 + θ)=cosθ </span>
<span>= Cos45°Cos30° - Sin45°Sin30° </span>
<span>Cos165° </span>
<span>= Cos ( 90° + 45°+30° ) </span>
<span>= -Sin( 45°+30° )..... (∵Cos(90 + θ)=-Sinθ </span>
<span>= Sin45°Cos30° + Cos45°Sin30° </span>
<span>Tan165° </span>
<span>= Tan ( 90° + 45°+30° ) </span>
<span>= -Cot( 45°+30° )..... (∵Cot(90 + θ)=-Tanθ </span>
<span>= -1/tan(45°+30°) </span>
<span>= -[1-tan45°.Tan30°]/[tan45°+Tan30°] </span>
<span>Substitute the above values with the following... These should be memorized </span>
<span>Sin 30° = 1/2 </span>
<span>Cos 30° =[Sqrt(3)]/2 </span>
<span>Tan 30° = 1/[Sqrt(3)] </span>
<span>Sin45°=Cos45°=1/[Sqrt(2)] </span>
<span>Tan 45° = 1</span>
I would go with B i had this question on my test and i got it correct
Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =
(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =
= 2a^3 - 2(a^3)/3 = [4/3](a^3)
Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = <span>a^3
ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =
Limit of </span><span><span>a^3 / {[4/3](a^3)} </span>as a -> 0 = 1 /(4/3) = 4/3
</span>
Step 1: Create an equation with a slope of 6
y=6x+b
Step 2: Substitute x and y by with the point (1,2) and solve the equation for b
y=6x+b
2=6(1)
2=6
2=6+b
b=-4
Step 3: Substitute -4 for b in the equation
y=6x+b
y=6x+(-4)
y=6x-4
The equation that has a slope of 6 and passes through the point (1,2) in point-slope form:
y=6x-4
