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D is on side BC of triangle ABC. We know $ABC \sim ACD$ and angle A = 49 degrees.  If BC = CD, what is angle B in degrees? 

 Jun 25, 2021
 #1
avatar+139 
+1

if $ \overset{-}{BC} = \overset{-}{CD}$ -- that doesnt make any sense! point D cuts $ \overset{-}{BC} $ so it just wouldnt work... unless i am missing somehting.

 Jun 25, 2021
 #2
avatar+121000 
+1

                  A=  49

 

B            D                C   

 

Agree  with  UTS.....if D is on BC, then    BC   cannot =  CD

 

 

cool cool cool                             

 Jun 25, 2021
 #3
avatar+1425 
+4

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.

 

 Jun 25, 2021
 #4
avatar+1425 
+3

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

jugoslav  Jun 26, 2021

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