Circular Measure + Trigonometry?

Question

okay again, i have two more questions. for fun.

Q1: A and B are two points on the circumference of a circle centre O and radius 4 cm. If the chord AB is of length 6cm. find the angle the minor arc AB subtends at the centre(in radians to 2 decimal places) and the area of the corresponding sector OAB.

Q2: A and B are two points on the circumference of a circle centre O and radius r. The minor arc AB subtends an angle θ radians at O. If the area of the minor segment bounded by the chord AB and the circle is one quarter of the area of the minor sector AOB, show that 4 sin θ = 3θ.

thanks again. =p

Answer 1

Here we go again. You ask good questions.

On Question 1, let C be the midpoint of AB. Then ACO is a right triangle; OC =sqrt(16 - 9) = sqrt(7).

sin AOC = 3/4 ==> angle AOC = 0.848 radians
angle AOB = 1.70 radians

The area of circle O is 16 pi, so the area of sector OAB is
(16 pi)[1.70/(2 pi)] = 8 times angle AOB = 13.57 sq cm


Question 2. Let t stand for theta, because I don't know how to make a "theta" on my computer, and let C be the midpoint of AB.

Area(sector) = (pi r^2)[t/(2 pi)] = (tr^2)/2
Area(AOB) = [r sin(t/2)] [ r cos(t/2)] = r^2 sin(t/2) cos(t/2)

I might have to use half-angle formulas here, which I'd have to look up, but we'll see ...

Area(segment) = Area(sector) - Area(AOB) = (1/4) Area(sector)
4 - 4 Area(AOB)/Area(sector) = 1
4 Area(AOB)/Area(sector) = 3

4 r^2 sin(t/2) cos(t/2) / [(tr^2)/2] = 3
8 sin(t/2) cos(t/2) = 3t

We're going to have to use the double angle formulas. Yuck! Let u = t/2, so t = 2u. According to the book,

sin(2u) = 2 sin u cos u

and substituting back,

sin t =2 sin(t/2) cos(t/2)

Hence, 4 sin t = 3t

Good questions!

Answer 2

1.let the central angl be 2x.sin x=3/4 and so x =48*35'=0.855 radians so 2x=1.71 radians
so the central angle 2x=1.71 radians
2.area of the sector=(theta/360)(pi*r^2)
and area of the triangle=1/2*r^2*sin theta
given (theta/360)*pi*r^2-1/2*r^2 sin theta=1/4*(theta/360)*pi*r^2
3/4 (theta/360)*pir^2=1/2*r^2*sin theta
3/8 theta=1/2 sin theta
therefore 3 theta=4 sin theta

Answer 3

You sound as whilst you're bearing on an already existent pastime wherein the employer did take such measures. And define, "such" measures. they have actual taken measures, and that i come across video games functional yet you're question is to imprecise to answer properly.

Answer 4

c^2 = a^2 + b^2 - 2abcos(theta)
6^2 = 4^2 + 4^2 - 2(4)(4)cos(theta)
36 = 32 - 32cos(theta)
Theta = acos(-4/32)
Theta is about 1.70 radians

Area = (base)(height) / 2
Area = (6 cm)(2.64575 cm) / 2
Area = 7.94 cm^2

Answer 5

I'm curious about the "for fun" part of your question.