Find the Area of triangle in the semi-circle
In the above figure, O is the centre of the circle.
If $angle BCO=30 ^circ$ and BC=$12 sqrt 3$, what is the area of triangle ABO?
I worked like OA=OB=OC(radii of the circle).
So, $angle OBC=30^circ,angle BOC=120^circ$
$angle AOB=60^circ,angle ABO=60^circ,angle OAB=60^circ$
Triangle AOB comes to be an equilateral triangle.
How Do I find OA?
Please help.
geometry
asked 2 days ago
user3767495
3348
add a comment |
In the above figure, O is the centre of the circle.
If $angle BCO=30 ^circ$ and BC=$12 sqrt 3$, what is the area of triangle ABO?
I worked like OA=OB=OC(radii of the circle).
So, $angle OBC=30^circ,angle BOC=120^circ$
$angle AOB=60^circ,angle ABO=60^circ,angle OAB=60^circ$
Triangle AOB comes to be an equilateral triangle.
How Do I find OA?
Please help.
geometry
asked 2 days ago
user3767495
3348
Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
That's the one.
– Mark Bennet
yesterday
add a comment |
In the above figure, O is the centre of the circle.
If $angle BCO=30 ^circ$ and BC=$12 sqrt 3$, what is the area of triangle ABO?
I worked like OA=OB=OC(radii of the circle).
So, $angle OBC=30^circ,angle BOC=120^circ$
$angle AOB=60^circ,angle ABO=60^circ,angle OAB=60^circ$
Triangle AOB comes to be an equilateral triangle.
How Do I find OA?
Please help.
geometry
asked 2 days ago
user3767495
3348
In the above figure, O is the centre of the circle.
If $angle BCO=30 ^circ$ and BC=$12 sqrt 3$, what is the area of triangle ABO?
I worked like OA=OB=OC(radii of the circle).
So, $angle OBC=30^circ,angle BOC=120^circ$
$angle AOB=60^circ,angle ABO=60^circ,angle OAB=60^circ$
Triangle AOB comes to be an equilateral triangle.
How Do I find OA?
Please help.
geometry
geometry
asked 2 days ago
user3767495
3348
asked 2 days ago
user3767495
3348
asked 2 days ago
user3767495
3348
asked 2 days ago
user3767495
3348
asked 2 days ago
user3767495
3348
3348
Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
That's the one.
– Mark Bennet
yesterday
add a comment |
Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
That's the one.
– Mark Bennet
yesterday
Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
That's the one.
– Mark Bennet
yesterday
That's the one.
– Mark Bennet
yesterday
add a comment |
2 Answers
2
active
oldest
votes
The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.
edited 2 days ago
amWhy
191k28224439
answered 2 days ago
Andarrkor
306
add a comment |
Let $AO=x$.
Thus, $AB=2x$ and since $measuredangle BCO=30^{circ},$ we obtain: $$AB=frac{1}{2}AC=x.$$
Now, by Pythagoras
$$AC^2-AB^2=BC^2$$ or
$$(2x)^2-x^2=(12sqrt3)^2,$$ which gives $$x=12.$$
Can you end it now?
answered yesterday
Michael Rozenberg
95.4k1588184
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.
edited 2 days ago
amWhy
191k28224439
answered 2 days ago
Andarrkor
306
add a comment |
The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.
edited 2 days ago
amWhy
191k28224439
answered 2 days ago
Andarrkor
306
add a comment |
The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.
edited 2 days ago
amWhy
191k28224439
answered 2 days ago
Andarrkor
306
The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.
edited 2 days ago
amWhy
191k28224439
answered 2 days ago
Andarrkor
306
edited 2 days ago
amWhy
191k28224439
edited 2 days ago
amWhy
191k28224439
edited 2 days ago
amWhy
191k28224439
191k28224439
answered 2 days ago
Andarrkor
306
answered 2 days ago
Andarrkor
306
answered 2 days ago
Andarrkor
306
306
add a comment |
add a comment |
Let $AO=x$.
Thus, $AB=2x$ and since $measuredangle BCO=30^{circ},$ we obtain: $$AB=frac{1}{2}AC=x.$$
Now, by Pythagoras
$$AC^2-AB^2=BC^2$$ or
$$(2x)^2-x^2=(12sqrt3)^2,$$ which gives $$x=12.$$
Can you end it now?
answered yesterday
Michael Rozenberg
95.4k1588184
add a comment |
Let $AO=x$.
Thus, $AB=2x$ and since $measuredangle BCO=30^{circ},$ we obtain: $$AB=frac{1}{2}AC=x.$$
Now, by Pythagoras
$$AC^2-AB^2=BC^2$$ or
$$(2x)^2-x^2=(12sqrt3)^2,$$ which gives $$x=12.$$
Can you end it now?
answered yesterday
Michael Rozenberg
95.4k1588184
add a comment |
Let $AO=x$.
Thus, $AB=2x$ and since $measuredangle BCO=30^{circ},$ we obtain: $$AB=frac{1}{2}AC=x.$$
Now, by Pythagoras
$$AC^2-AB^2=BC^2$$ or
$$(2x)^2-x^2=(12sqrt3)^2,$$ which gives $$x=12.$$
Can you end it now?
answered yesterday
Michael Rozenberg
95.4k1588184
Let $AO=x$.
Thus, $AB=2x$ and since $measuredangle BCO=30^{circ},$ we obtain: $$AB=frac{1}{2}AC=x.$$
Now, by Pythagoras
$$AC^2-AB^2=BC^2$$ or
$$(2x)^2-x^2=(12sqrt3)^2,$$ which gives $$x=12.$$
Can you end it now?
answered yesterday
Michael Rozenberg
95.4k1588184
answered yesterday
Michael Rozenberg
95.4k1588184
answered yesterday
Michael Rozenberg
95.4k1588184
answered yesterday
Michael Rozenberg
95.4k1588184
95.4k1588184
add a comment |
add a comment |
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Can you find a right-angled triangle and use Pythagoras?
– Mark Bennet
2 days ago
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
– user3767495
2 days ago
That's the one.
– Mark Bennet
yesterday