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# TRIGONOMETRY

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In right triangle ABC, we have angle BAC = 90 degrees and D is on AC such that BD bisects ABC. If AB = 12 and BC = 15, then what is the cosine of angle BDC?

Dec 26, 2019

#1
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Draw the triangle ABC, Plot D on AC and draw a line from angle ABC to point D, Thus, ABD=DBC

In the right angle triangle ABC, we know that AB=12, BC=15 therefore, AC=9

We have all sides so we can use either sine or cosine or tangent to find angle ABC

let's use cosine

Let, angle ABC=y

cos(y)=12/15

y=cos^-1(12/15) = 36.869897645844

Since BD bisects ABC

Therefore BDC=BDA as we said earlier and we know that DBC+DBA=ABC thus, BDC= 1/2ABC

so BDC=(1/2)36.869897645844 = 18.434948822922

We know 2 angles in a triangle so the third angle

180-(36.869897645844+90)=53.130102354156 (Angle BCA)

In triangle BDC, we know 2 angles which are Angle DBC=18.434948822922 and angle BCA=53.1301023541456

Therefore angle BDC=180-(53.1301023541456+18.434948822922) = 108.4349488229324

cos(108.4349488229324) = -0.316227766017

Note finding angle BDC is easier if we just added angle BAD+DBA since BDC is exterior angle of triangle ABD.

Dec 26, 2019
#2
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See the image AB = 12       BC  = 15       so     AC =   9

Since BD bisects   angle ABC,  then

AD / DC =  AB/ BC

So   AD =  (4/9)(9) =  4

So    D  =(4,0)    and  DC =  AC - AD  =  9 - 4    =  5

And the distance from B to D  =  sqrt  (12^2  + 4^2 )  =  sqrt  (160)

So....using the Law of Cosines....

BC^2  =  DC^2  + BD^2   -2(DC* BD)cos(BDC)

15^2  = 5^2  +  160  -  2 ( 5sqrt(160)) cos(BDC)

40 =  -10sqrt (160) cos (BDC)

-40 /  (10 sqrt (160) )  =  cos (BDC)

-4 / [ sqrt (160) ]  =cos (BDC)

-4sqrt (160)  / 160  = cos (BDC)

-sqrt (160) / 40  = cos (BDC)

- sqrt (10)  / 10  =  cos (BDC)   Dec 26, 2019