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O is the center of the semicircle. If angle BCO=30 degrees and BC = \[6\sqrt{3}\] what is the area of triangle ABO?
|dw:1435335512997:dw|
Do you know how to find the area of a triangle?

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Other answers:

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

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