In triangle ABC, $\angle$ A equals 120 degrees. A point D is inside the triangle such that $\angle$DBC = 2 $\times \angle $ABD and $\angle$DCB = 2 $\times \angle$ACD. Determine the measure, in degrees, of $\angle$ BDC. [asy][asy] pair A = (5,4); pair B = (0,0); pair C = (10,0); pair D = (5,2.5) ; draw(A--B); draw(B--C); draw(C--A); draw (B--D--C); label ("A", A, dir(45)); label ("B", B, dir(45)); label ("C", C, dir(45)); label ("D", D, dir(45)); [/asy][/asy]
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08.06.2017 05:44
08.06.2017 07:34
Isn't it 140 degrees? They are not angle bisectors. $180-2/3\cdot60=180-40=140$
08.06.2017 08:14
Idk i got 150
28.08.2019 09:04
Let $\angle{ABD}=a, \text{and let } \angle {ACD}=b$ Therefore, $\angle{ DBA}=2a$$ \text{ and } $$\: \angle{ACD}=2b$ Therefore, $180=120+a+2a+b+2b$ $\rightarrow 180=120+3a+3b$ $\rightarrow 60=3a+3b$ $\rightarrow 20=a+b$ Now, we know that the sum of the angles of triangle $DBC=180,$ so $180=2a+2b+\angle{D}$ We substitute a+b for 20, and we get $180=40+\angle {D}$. $\boxed{\angle {D}=140 \text{ degrees}}$
28.08.2019 19:39
bruh why the 2 year bump
29.08.2019 07:20
why not?
29.08.2019 20:20
because the topic has already been dealt with and bumping old threads will just clog up the forum