# Thread: Arc length and sector area

1. ## Arc length and sector area

Dear Math Help Forum members,

This question has been niggling me for an hour. I couldn't solve it and I am in a dire situation. Please give me advice as to how to solve this question (don't overlook the parts). I would like to have your comments as soon as possible. All I know is that for part (a), the answer is 2 x 1.3rad. = 2.6rad.

Thank you so much and have a nice day.

2. ## Re: Arc length and sector area

Draw lines from the circle centre "O" to "A", "B" and "C".
Label the lengths |AB| and |AC| and the lengths |OB|, |OA| and |OC|.

Consider the triangles OAB and OAC.
Are they congruent or not ?
Then try to find some internal angles in the triangles,
allowing you to find |AB|

3. ## Re: Arc length and sector area

Originally Posted by PythagorasNeophyte
Dear Math Help Forum members,

This question has been niggling me for an hour. I couldn't solve it and I am in a dire situation. Please give me advice as to how to solve this question (don't overlook the parts). I would like to have your comments as soon as possible. All I know is that for part (a), the answer is 2 x 1.3rad. = 2.6rad.

Thank you so much and have a nice day.
For a) The angle made from two points to the centre is always double the angle made from three points on the circle.

4. ## Re: Arc length and sector area

Hello pythagorasneophyte,
Draw in lines BC,BO,AO extended to arcBC (partial diameter of circle r=20 and radius of circle = AB),OD and OE the perpendicular bisectors of AB and BC

BAC =1.3 rad = 74.5 deg BOC = 149deg BAO = 37,24 deg
cosBAO = 0.796 = 1/2 AB/20 AB = 31.8

In like manner BC = 38.5

Calculate the area of two segments BQC and BPC. Crescent is equal to the larger segment minus the smaller.

bjh

5. ## Re: Arc length and sector area

I'm so sorry. I couldn't find the length AB, despite knowing that triangles BOA and COA are congruent triangles, and that each is an isosceles triangle.

In triangle BOA, BO = OA = 20cm because they are the base sides of an isosceles triangle. This triangle, however, is not right-angled, so I can't use the trigonometrical ratios for finding AB.

Can you show me how to do it?

6. ## Re: Arc length and sector area

Originally Posted by PythagorasNeophyte
I'm so sorry. I couldn't find the length AB, despite knowing that triangles BOA and COA are congruent triangles, and that each is an isosceles triangle.

In triangle BOA, BO = OA = 20cm because they are the base sides of an isosceles triangle. This triangle, however, is not right-angled, so I can't use the trigonometrical ratios for finding AB.

Can you show me how to do it?
You may use the fact that angle BAC is 1.3 radians
and angle OAC equals angle OAB and so also equals angle OBA.

Hence angle AOB is $\displaystyle \pi-1.3$ radians.

Write the value of angle OBA or OAB.

You may now use the Sine Rule to calculate |AB|.

Alternatively, the angles AOB and COB are equal, so either is

$\displaystyle \frac{2\pi-2.6}{2}$ radians, using the angle at the centre.

Then obtain angle OBA or OAB using the fact that the triangle is isosceles
and all 3 angles in a triangle sum to $\displaystyle \pi$ radians.

7. ## Re: Arc length and sector area

Hi pythagorasneophyte,
Following my previous instructions triangle ODA is the right triangle in which to use the cos ( adjacent over hyp).The angleBAO is 37.24 deg and AB =31.8 .The adj leg is 1/2 AB

bjh

8. ## Re: Arc length and sector area

Part a) has been answered its 2.6 rad. (Angle at center = 2* angle at circumference)

Part b)
Using Sine law in $\displaystyle \Delta$OBC we have
BC/sin(2.6) = r/sin(1.3) - Eq1
Using Sine law in $\displaystyle \Delta$ABC we have
BC/sin(1.3) = R /sin(.65) - Eq2
From Eq1 & Eq2 we have
$\displaystyle R = \frac{r.sin(.65).sin(2.6)}{(sin(1.3))^2}$

Part c)
Perimeter of crescent is arc(BQC) + arc(BPC) = $\displaystyle r.(2.6)+R.(1.3)$

Part d)
area of crescent = $\displaystyle (area(sector BQOC) - \Delta BOC ) - (area(sector BPOC) - \Delta ABC)$ = $\displaystyle \frac{1}{2}((r^2(2.6) - r^2.sin(2.6))- (R^2(1.3) - R^2.sin(1.3)))$

Kalyan.
Note: All angles are in radians.