Find the Area of triangle in the semi-circle












0












$begingroup$


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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$endgroup$












  • $begingroup$
    Can you find a right-angled triangle and use Pythagoras?
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 2:01










  • $begingroup$
    You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
    $endgroup$
    – user3767495
    Dec 25 '18 at 2:02










  • $begingroup$
    That's the one.
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 8:10
















0












$begingroup$


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









$endgroup$












  • $begingroup$
    Can you find a right-angled triangle and use Pythagoras?
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 2:01










  • $begingroup$
    You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
    $endgroup$
    – user3767495
    Dec 25 '18 at 2:02










  • $begingroup$
    That's the one.
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 8:10














0












0








0





$begingroup$


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









$endgroup$




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 Dec 25 '18 at 1:53









user3767495user3767495

4078




4078












  • $begingroup$
    Can you find a right-angled triangle and use Pythagoras?
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 2:01










  • $begingroup$
    You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
    $endgroup$
    – user3767495
    Dec 25 '18 at 2:02










  • $begingroup$
    That's the one.
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 8:10


















  • $begingroup$
    Can you find a right-angled triangle and use Pythagoras?
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 2:01










  • $begingroup$
    You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
    $endgroup$
    – user3767495
    Dec 25 '18 at 2:02










  • $begingroup$
    That's the one.
    $endgroup$
    – Mark Bennet
    Dec 25 '18 at 8:10
















$begingroup$
Can you find a right-angled triangle and use Pythagoras?
$endgroup$
– Mark Bennet
Dec 25 '18 at 2:01




$begingroup$
Can you find a right-angled triangle and use Pythagoras?
$endgroup$
– Mark Bennet
Dec 25 '18 at 2:01












$begingroup$
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
$endgroup$
– user3767495
Dec 25 '18 at 2:02




$begingroup$
You mean AC=2r,AB=r and $angle ABC=90^circ$ ?
$endgroup$
– user3767495
Dec 25 '18 at 2:02












$begingroup$
That's the one.
$endgroup$
– Mark Bennet
Dec 25 '18 at 8:10




$begingroup$
That's the one.
$endgroup$
– Mark Bennet
Dec 25 '18 at 8:10










2 Answers
2






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$begingroup$

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











$endgroup$





















    0












    $begingroup$

    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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      active

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      2 Answers
      2






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      active

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      2












      $begingroup$

      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











      $endgroup$


















        2












        $begingroup$

        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











        $endgroup$
















          2












          2








          2





          $begingroup$

          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











          $endgroup$



          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 Dec 25 '18 at 3:00









          Namaste

          1




          1










          answered Dec 25 '18 at 2:14









          AndarrkorAndarrkor

          496




          496























              0












              $begingroup$

              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









              $endgroup$


















                0












                $begingroup$

                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









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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









                  $endgroup$



                  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 Dec 25 '18 at 6:33









                  Michael RozenbergMichael Rozenberg

                  107k1895199




                  107k1895199






























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