Find the Area of triangle in the semi-circle












0














enter image description here



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







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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
















0














enter image description here



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







share|cite|improve this question






















  • 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














0












0








0







enter image description here



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







share|cite|improve this question













enter image description here



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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










2 Answers
2






active

oldest

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2














The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.






share|cite|improve this answer































    0














    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?






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






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      active

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      active

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      2














      The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.






      share|cite|improve this answer




























        2














        The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.






        share|cite|improve this answer


























          2












          2








          2






          The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.






          share|cite|improve this answer














          The ABC angle is a right angle, so $12sqrt{3} tan(30) = AB$ then $frac{ABcdot BC}{2}$ and you got it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago









          amWhy

          191k28224439




          191k28224439










          answered 2 days ago









          Andarrkor

          306




          306























              0














              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?






              share|cite|improve this answer


























                0














                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?






                share|cite|improve this answer
























                  0












                  0








                  0






                  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?






                  share|cite|improve this answer












                  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?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Michael Rozenberg

                  95.4k1588184




                  95.4k1588184






























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