anonymous
  • anonymous
Please Help!!!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
O is the center of the semicircle. If angle BCO=30 degrees and BC = \[6\sqrt{3}\] what is the area of triangle ABO?
anonymous
  • anonymous
|dw:1435335512997:dw|
misssunshinexxoxo
  • misssunshinexxoxo
Do you know how to find the area of a triangle?

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More answers

anonymous
  • anonymous
yes
anonymous
  • anonymous
bh/2
anonymous
  • anonymous
help someone...
anonymous
  • anonymous
@misty1212
mathstudent55
  • mathstudent55
|dw:1435338018836:dw|
mathstudent55
  • mathstudent55
|dw:1435338107837:dw|
mathstudent55
  • mathstudent55
Do you understand so far?
anonymous
  • anonymous
wait how do we know it's 60?
anonymous
  • anonymous
what about the triangle on the right side what do those angles equal to?
mathstudent55
  • mathstudent55
I'll show you each angle I found, one at a time, and will explain.
anonymous
  • anonymous
okay
mathstudent55
  • mathstudent55
Start here. We have the side that is \(6 \sqrt 3\), the angle of 60 deg, and all three radii are congruent. |dw:1435338257728:dw|
anonymous
  • anonymous
okay
mathstudent55
  • mathstudent55
Now look at angle ABC. It is an inscribed angle. An inscribed angle is half the measure of the intercepted arc. Arc AC (on the bottom side) is 180 deg because it is a semicircle since we have the diameter AC.
mathstudent55
  • mathstudent55
That makes angle ABC 90 degrees, and triangle ABC a right triangle.
mathstudent55
  • mathstudent55
|dw:1435338496200:dw|
anonymous
  • anonymous
right i remember all inscribed angles are 90 degrees
mathstudent55
  • mathstudent55
Now look at triangle ABC. One angle is 90 deg. One angle is 30 deg. That means the left angle, angle A is 60 deg.
mathstudent55
  • mathstudent55
|dw:1435338551348:dw|
anonymous
  • anonymous
30 60 90 yea
mathstudent55
  • mathstudent55
Correct.
mathstudent55
  • mathstudent55
Now look at the congruent sides of triangle AOB.
mathstudent55
  • mathstudent55
|dw:1435338598897:dw| Triangle AOB is isosceles. Opp angles to the congruent sides are congruent.
mathstudent55
  • mathstudent55
|dw:1435338640262:dw|
mathstudent55
  • mathstudent55
Once you have two angles of triangle AOB are 60 deg each, the third angle must also be 60 deg, and it's an equilateral triangle.
anonymous
  • anonymous
sorry i lost you
mathstudent55
  • mathstudent55
|dw:1435338690645:dw|
anonymous
  • anonymous
Triangle AOB is isosceles. Opp angles to the congruent sides are congruent.
anonymous
  • anonymous
what do u mean by that
mathstudent55
  • mathstudent55
Did you understand that angle A of triangle AOB is 60 deg?
anonymous
  • anonymous
yes
mathstudent55
  • mathstudent55
There is a theorem that states: If a triangle has two congruent sides, then the angles opposite those sides are congruent.
mathstudent55
  • mathstudent55
That means, in an isosceles triangle, the base angles are congruent.
anonymous
  • anonymous
hmmmm.... okay
mathstudent55
  • mathstudent55
|dw:1435338811305:dw|
anonymous
  • anonymous
yes angles B and C are equal
anonymous
  • anonymous
but the A is different....
anonymous
  • anonymous
so how does it relate
mathstudent55
  • mathstudent55
Ok. Now back to our problem. Notice that triangle AOB has two radii as sides, AO and OB. All radii in a circle are congruent, so sides AO and OB are congruent sides in a triangle.
mathstudent55
  • mathstudent55
That means the base angles, BAO and ABO are congruent angles.
mathstudent55
  • mathstudent55
|dw:1435339099484:dw|
mathstudent55
  • mathstudent55
We now know that angle ABO also measures 60 deg. |dw:1435339121331:dw|
mathstudent55
  • mathstudent55
Once two angles of a triangle are known, you can find the third angle. 60 + 60 + m
mathstudent55
  • mathstudent55
Now we know that triangle AOB is also equilateral in addition to being isosceles.
anonymous
  • anonymous
Cant it be this? |dw:1435339216515:dw|
anonymous
  • anonymous
then it wouldn't be 60 60 60
mathstudent55
  • mathstudent55
According to the given info, it must be equilateral.
anonymous
  • anonymous
okay thank you
anonymous
  • anonymous
im gonna have to look over this a few times im still having a hard time understanding why its all 60
mathstudent55
  • mathstudent55
This is not the end of the problem. We still need to find the area of triangle ABO.
mathstudent55
  • mathstudent55
Triangle ABC is a 30-60-90 triangle. AB is the short leg, and BC is the long leg. From the ratio of the lengths of the sides of a 30-60-90 triangle, we have: \(1~:~ \sqrt 3~:~ 2\) for the ratio of short leg : long leg : hypotenuse
mathstudent55
  • mathstudent55
We see that the long leg is \(\sqrt 3\) times the length of the short leg. That means the short leg is \(\sqrt 3\) times shorter than the long leg.
mathstudent55
  • mathstudent55
For triangle ABC, the short leg is AB, and the long leg is BC. Since AB is \(\sqrt 3\) times shorter than BC, AB is 6 units long.
mathstudent55
  • mathstudent55
|dw:1435339904347:dw|
mathstudent55
  • mathstudent55
Now the problem is simply to find the area of an equilateral triangle whose side has a length of 6.
anonymous
  • anonymous
so bh/2 right?
mathstudent55
  • mathstudent55
Yes, but we need to find the height. Let's just look at the triangle now.
anonymous
  • anonymous
6 would be the opposite of 30 degrees
anonymous
  • anonymous
and height is given as \[6\sqrt{3}\]
mathstudent55
  • mathstudent55
|dw:1435339974085:dw|
anonymous
  • anonymous
ohhhhh yeaaaa
mathstudent55
  • mathstudent55
|dw:1435340010942:dw|
mathstudent55
  • mathstudent55
Inside triangle AOB, we have two 30-60-90 triangles. |dw:1435340060713:dw|
mathstudent55
  • mathstudent55
XO is 3, since 3 is half of 6. Now we need h, the height of triangle AOB. h is also the long leg of the 30-60-90 triangle BXO.
anonymous
  • anonymous
so it would be \[9\sqrt{3}\]
mathstudent55
  • mathstudent55
Remember, the long leg is \(\sqrt 3\) times longer than the short leg. |dw:1435340159438:dw|
mathstudent55
  • mathstudent55
|dw:1435340236323:dw|
anonymous
  • anonymous
i understand everything except how it's a 60-60-60- triangle
mathstudent55
  • mathstudent55
Now we can find the area of triangle AOB: \(A = \dfrac{bh}{2} \) \(A = \dfrac{6 \times 3 \sqrt 3}{2} \) \(A = 9\sqrt 3\)
mathstudent55
  • mathstudent55
You are correct.
mathstudent55
  • mathstudent55
I'll show you that again, about the 60-60-60 triangle.
anonymous
  • anonymous
okay
mathstudent55
  • mathstudent55
You understood up to here that angle ABC is a right angle, and that angle BAC must be 60 deg.? |dw:1435340411852:dw|
anonymous
  • anonymous
yes
mathstudent55
  • mathstudent55
Now let's mark the congruent segments we have because they are radii of the circle. |dw:1435340566038:dw|
mathstudent55
  • mathstudent55
Ok?
anonymous
  • anonymous
yes yes
anonymous
  • anonymous
=)
mathstudent55
  • mathstudent55
Great. Now let's concentrate only on triangle AOB.
mathstudent55
  • mathstudent55
|dw:1435340649665:dw|
anonymous
  • anonymous
so AO and BO makes it an isosceles
mathstudent55
  • mathstudent55
We have a triangle, ABO, in the figure above. Sides AO and BO are congruent. That means the opposite angles, angles A and B are congruent.
mathstudent55
  • mathstudent55
Yes, triangle AOB is isosceles.
mathstudent55
  • mathstudent55
|dw:1435340749447:dw|
anonymous
  • anonymous
but we dont now what AB is it could be anything
mathstudent55
  • mathstudent55
Yes, so far you are correct. We don;t know anything about AB.
anonymous
  • anonymous
oh noooooooooooooooooooooooo so because AO is 60 then BO would be 60 so the remaining would also be 60 .....
mathstudent55
  • mathstudent55
We already know that angle A measures 60, so we can conclude that angle B also measures 60. That is because angles A and B are opposite congruent sides,so they must be congruent.
mathstudent55
  • mathstudent55
|dw:1435340885281:dw|
anonymous
  • anonymous
i get it now ohhhhhhh okay wow i'm dumb that took me forever to get ...
mathstudent55
  • mathstudent55
\(\Huge \bf \color{red}{BINGO!!!} \)
anonymous
  • anonymous
Thank you so much <333333333333333333333333333333
mathstudent55
  • mathstudent55
You are very welcome. I'm glad we took the extra time for you to understand it completely.
anonymous
  • anonymous
Couldn't have done it without you thanks!!!!!! =*)
mathstudent55
  • mathstudent55
You're welcome. Glad to help.

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