# In triangle ABC, angle ABC=15 degrees, angle ACB=30 degrees, AD is the median to side BC, A-D-C. angle BAD=?

**Answer**

We will use the Sine Rule.

In triangle ABD,

AD / sin(ABD) = BD / sin(BAD)

So

AD / BD = sin(ABD) / sin(BAD)

In triangle ADC,

AD / sin(ACD) = CD / sin(CAD)

and AD / CD = sin(ACD) / sin(CAD)

Since BD = CD, ( because D is median in BC ),

implies AD/BD = AD/CD, thus

sin(ABD) / sin(BAD) = sin(ACD) / sin(CAD)

and also BAC = 180 – ABC – ACB = 180 – 15 – 30 = 135.

sin ( ACD ) = sin (30) = 0.5

sin (ABD ) = sin (15) = 0.2588

This yields in the above,

sin(ABD) / sin(BAD) = sin(ACD) / sin(CAD) becomes,

sin(CAD)/sin(BAD) = sin(ACD) / sin(ABD)

sin(CAD)/sin(BAD) = 0.5 / 0.2588 = 1.932 ( roughly ). .. ..(1)

Also,

sin(CAD) = sin ( BAC – BAD )

sin(CAD) = sin ( 135 – BAD )

sin(CAD) = sin(135) cos(BAD) – cos(135) sin ( BAD )

Now, sin ( 135 ) = 0.7071 and cos 135 = – 0.7071. so,

sin(CAD) = sin(135) cos(BAD) – cos(135) sin ( BAD )

sin(CAD) = ( 0.7071) cos(BAD) + ( 0.7071 ) sin ( BAD )

sin(CAD) = ( 0.7071) cos(BAD) + ( 0.7071 ) sin ( BAD ) … (2)

Put (2) into (1) to get ,

[( 0.7071) cos(BAD) + ( 0.7071 ) sin ( BAD )] /sin(BAD)

= 1.932 , That is,

cos(BAD) + sin (BAD) = 2.7323 sin(BAD)

That is,

1.7323 sin(BAD) = cos(BAD)

sin(BAD) / cos(BAD) = 1.7323

Tan (BAD) = 1.7323 [ because sinx / cosx = tan x ]

BAD = Tan^(-1) 1.7323 = 60 degrees ( approximately ).