Find the m∠BCA, if m∠DCE = 70° and m∠EDC = 50°.
50°
60°
70°
80°
![Find the mBCA if mDCE 70 and mEDC 50 50 60 70 80 class=](https://us-static.z-dn.net/files/db3/18bb22ff6da3f338ea70fd915714eef6.png)
Answer-
[tex]\boxed{\boxed{m\angle BCA=60^{\circ}}}[/tex]
Solution-
Given here,
As [tex]BC||DE[/tex], and DC is the transversal, so
[tex]\Rightarrow \angle EDC=\angle DCB\ \ \ (\because \text{Alternate Interior Angle})[/tex]
[tex]\Rightarrow m\angle DCB=50^{\circ}[/tex]
Ans also [tex]\angle DCE,\ \angle DCB,\ \angle BCA[/tex] are complementary angles. So
[tex]\Rightarrow m\angle DCE+ m\angle DCB+m\angle BCA=180^{\circ}[/tex]
[tex]\Rightarrow m\angle BCA=180^{\circ}-m\angle DCE-m\angle DCB[/tex]
[tex]\Rightarrow m\angle BCA=180^{\circ}-70^{\circ}-50^{\circ}=60^{\circ}[/tex]