Asked by | 4th Mar, 2009, 09:34: PM
Consider an equilateral triangle ABC with each side = a(say).
So all the angles of the triangle must be 60 degrees each.
Draw AM perpendicular from A on BC.
In triangle AMB and AMC,
angle AMB= angle AMC ( each is 90 degrees)
AB=AC ( sides of equilateral triangle)
AM=AM ( common)
triangle AMB congruent to triangle AMC(RHS)
Angle BAM= angle CAM(C.P.C.T.)
CM=BM(C.P.C.T.), each of CM and BM will be equal to half of BC =a/2
So, each must be 30 degrees as the full angle A is 60 degrees.
Consider triangle AMB( say) you can take triangle AMC also.
sin MAB=sin 30=opposite side/ hypotenuse=(a/2) divided by a=1/2
Thus we have proved that sin 30 =1/2
Do this question in the exam with the appropriate drawing as explained in the solution.
All the best!
Answered by | 4th Mar, 2009, 10:40: PM
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