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probable ..............................
Asked by | 04 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,

we have,

angle AMB= angle AMC ( each is 90 degrees)

AB=AC ( sides of equilateral triangle)

AM=AM ( common)

So

triangle AMB congruent to triangle AMC(RHS)

Angle BAM= angle CAM(C.P.C.T.)

and

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.

angle MAB=30

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 | 04 Mar, 2009, 10:40: PM

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