Sector areas.

RickyOswaldIOW · 2006-07-14 23:51:36

I have the following triangle from which I have to work out the area of the shaded part.

I am firstly working out the area of the triangle:
triangle area = (c*b)/2 = (5*12)/2 = 30

I know the formula to work out the area of the sector is:
(r²θ)/2

To work out θ I use:
a² = b² + c² -2bcCosθ

I can work out the length of a using:
a² = b² + c² -> a² = 12² + 5² -> a² = 169 ->
a = √169 = 13

Thus:
c² = b² + a² -2baCosθ
5² = 12² + 13² - 2(12)(13)Cosθ
25 = 313 - 312Cosθ
312Cosθ = 288
Cosθ = 288/312 = 0.9230...
θ = 0.4

And so the area of the segment is:
(r²θ)/2 = (12² * 0.4)/2 = 28.8

If I subtract this value from the area of the triangle I should be left with the area of the shaded part:
30 - 28.8 = 1.2

My book states the answer as 1.58?!

Last edited by rickyoswaldiow (2006-07-16 22:13:22)

John E. Franklin · 2006-07-15 05:33:42

I forgot the trig formula, so I don't know which statement is right.
One statement you said has 3 variables and another only 2 variables:
= b² + c² -2abCosθ (this line has b, c, a, and b again)

= b² + a² -2baCosθ (this line has b, a, and b and a again)
You probably did it all right, am I just pointing out a typo?

Last edited by John E. Franklin (2006-07-15 05:34:50)

DASET · 2006-07-15 07:24:33

the book answer is correct.

the area of the triangle is 30 as you said. your formula for the area of a sector is correct, but the value of your angle is not accurate. your method is very indirect, you choose a round about way of computing the angle and lost accuracy due to the lack of precision in your conversion to radians.

from your figure, the arctangent of 5/12 yields 22.62 degrees or 0.394791 radians. thus using the angle in radians, the area of the sector is 28.42 and the shaded area is 1.58 rounded to two digits after the decimal point.

RickyOswaldIOW · 2006-07-16 22:12:41

One statement you said has 3 variables and another only 2 variables

a² = b² + c² -2abCosθ is the genral formula but since my triangle was already labled I just had to rename the variables. it could have been x, y, z or anything really.

ou choose a round about way of computing the angle and lost accuracy due to the lack of precision in your conversion to radians

Do you mean in the step where I write "θ = 0.4"? I probably should not have rounded this figure up to 0.4, I'll retry the sum without rounding the value of theta.

DASET · 2006-07-16 23:18:34

...Do you mean in the step where I write "θ = 0.4"? I probably should not have rounded this figure up to 0.4, I'll retry the sum without rounding the value of theta.

yes. remember, whenever you have to subtract a number from another number which is nearly equal to it or whenever you expect the difference to be very small, you should carry out your calculations with as many significant digits as necessary to get an accurate answer.

another issue is your decision to use the law of cosines to find the angle theta. this approach required you to solve for the side c. too much unnecessary work. you were given a right triangle which means one of the angles is equal to 90 degrees. you were given the lengths of two sides of a right triangle. the easiest approach to finding the angle theta is by using the inverse (or arc) tangent function on your calculator, after dividing 5 by 12.

in an exam setting, you would have lost too much time using your approach. in the future, evaluate what is given in a problem and what is required, before you try to solve it.

cheers.

Last edited by DASET (2006-07-16 23:25:05)

RickyOswaldIOW · 2006-07-17 04:02:25

I have not been taught this method yet but it may appear soon. Thanks a lot for the help though

RickyOswaldIOW · 2006-07-17 08:08:45

Welcome to the forum DASET

Math Is Fun Forum

#1 2006-07-14 23:51:36

Sector areas.

#2 2006-07-15 05:33:42

Re: Sector areas.

#3 2006-07-15 07:24:33

Re: Sector areas.

#4 2006-07-16 22:12:41

Re: Sector areas.

#5 2006-07-16 23:18:34

Re: Sector areas.

#6 2006-07-17 04:02:25

Re: Sector areas.

#7 2006-07-17 08:08:45

Re: Sector areas.

Board footer