Thread: Pendulum

Hi guys, I have no clue to this question. Any help is appreciated. Ty.

The length of a pendulum is 20cm. If the tip of the pendulum swings through an angle of 86 degrees, find :

a) The arc length ABC (the distance through which the tip travels)
b) The area of triangle OAC.
c) The area of the sector.
d) The area of the minor segment.
e) The length BX.

Could you please explain what are the points A,B,C,O,X?

As for the first question, the arc length S of a circular sector, is connected to the radius r and the epicentric angle a (measured in radians), by the formula S=ra.

The length l of the chord (BC, if I am guessing the points right)
is l=2rsin(a/2).

And for the area E of the circlular segment, we have E=(1/2)ar^2.

These are enough to calculate all else

You mentioned all those letters but you did not explain them, or at least showed a figure re their locations.

If you cannot show a figure/picture, let me guess where/what these letters are, and I will give answers based on these guesses.

Based on the posted question,
>>>O is the center of the circle where the sector is a part of. O is where the pendulum is hinged.
>>>ABC is the circular arc defined by the tip of the pendulum. ABC is the arc of the sector.
>>>OA, OB, and OC are radii of the cicle or sector, and each one of them is 20cm long.
>>>OB is actually OXB and it is vertical. OB bisects central angle 86 degrees. OB bisects the chord AC or AXC at X. OB bisects the arc ABC at B.
>>>OABC is the sector; OAXC is the triangle; AxCB is the circular segment.
>>>BX is the "height of the segment" or the maximum distance between the chord and the arc.

---------------------------
a) The arc length ABC (the distance through which the tip travels)
arc = (radius)(central angle in radians)
arc ABC = 20[86*(pi/180)] = 30 cm -----answer.

b) The area of triangle OAC.
We use: Area of triangle = (1/2)(one side)(other side)*sin(included angle)
triangle Area = (1/2)(20)(20)sin(86deg) = 199.5 sq.cm. ----answer.

c) The area of the sector.
We use: Area of sector = (1/2)(radius)(arc)
sector Area = (1/2)(20)(30) = 300 sq.cm. -------answer.

d) The area of the minor segment.
Area of circular segment = (sector area) minus (triangle area)
segment Area = 300 -199.5 = 100.5 sq.cm. --------answer.

e) The length BX.
BX = OXB minus OX ----(i)
where
OXB = radius = 20cm.
OX is the altitude of triangle OAC based on the chord AXC.

In right triangle OXA,
OX = OA*cos(angle AOX) -----angle AOX = 86/2 = 43deg

Hence, substitutions in (i),
BX = 20 -20cos(43deg) = 20 -14.627 = 5.373 cm -------answer.