D is on side BC of triangle ABC. We know ABC∼ACD and angle A = 49 degrees. If BC = CD, what is angle B in degrees?
if −BC=−CD -- that doesnt make any sense! point D cuts −BC so it just wouldnt work... unless i am missing somehting.
A= 49
B D C
Agree with UTS.....if D is on BC, then BC cannot = CD
D is on side BC of triangle ABC. We know ABC ~ ACD and angle A = 49 degrees. If BD = CD, what is angle B in degrees?
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After a small correction, this becomes a very challenging question.
ΔABC ~ ΔACD BD = CD = 2 ∠B = ∠CAB = β
CD / AC = AC / BC
2 * 4 = (AC)2
AC = √8
sinβ / 2 = sin49 / √8
Angle B ≈ 32.253º
+175 
