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