All angles in the equilateral triangle are 60°
In the isosceles triangle the two base angles are
73 ° each ......... ( (180-34)/2 )
so angle ABD = 60+73 or 133°
Equilateral triangle ABC and isosceles triangle DBC share side BC. If angle BDC= 34 and BD=BC, what is the measure of angle ABD?
I don't know how to solve this questions, can someone guide me through?
3 answers
Start by drawing a diagram.
If BC=BD, then
∠BDC=∠BCD=34° (angles opposite equal sides.
Hence ∠CBD=180-(34+34)=112°
∠ABD=∠ABC+∠CBD=?
Recall that ΔABC is equilateral.
If BC=BD, then
∠BDC=∠BCD=34° (angles opposite equal sides.
Hence ∠CBD=180-(34+34)=112°
∠ABD=∠ABC+∠CBD=?
Recall that ΔABC is equilateral.
Go with MathMate, I misread the question