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ABC triangle. sin A=1/2, sin B=1 and BC=4. Calculate the surface of the ABC triangle.

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thales | (Level 1) Honors

Posted April 21, 2009 at 10:42 PM via web

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ABC triangle. sin A=1/2, sin B=1 and BC=4. Calculate the surface of the ABC triangle.

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neela | High School Teacher | (Level 3) Valedictorian

Posted May 25, 2009 at 4:47 PM (Answer #1)

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In triangle ABC,  SinB =1 .Therefore  B= 90 degree. The triangle is right angled at B and  AC  hypotenuse.

Sin A = 1/2 . Therefore  A = 30 degree and the remaing angle C = 180-(90+30) = 60 degree.

Therefore, BC  = AC sin A .  AC = BC/sinA = 4/(1/2) = 8

AB = ACcosA=8*cos30 = 8(sqrt3/2) = 4sqrt3

Therefore area of the triangle = (1/2)AB*BC

= (1/2)(4sqrt3)(4) = 8sqrt3 = (8sqrt3) = 13.8564 square units approximately.

 

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fahad-alfahad | eNotes Newbie

Posted June 5, 2009 at 5:31 PM (Answer #2)

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because of sin(A)=1/2 ,A=30 ,sin(B)=1,B=90

therefore ABC right angle triangle (90 , 60 , 30 )

so the side opposite  to the 30degree=1/2(chord)=4

chord=8 ======>the theird side =sqrt((8^2)-(4^2))=sqrt(48)

the surface of the ABC triangle=1/2 * 4 *sqrt(48)

=13.8564

 

 

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