If in the given figure, ∠ACB + ∠BAC = 80°, ∠BDE = 35°, ∠BCE = 45°, then the marked angle ∠CED is:

A. 150°
B. 120°
C. 160°
D. 135°
Answer: Option C
Solution (By Examveda Team)

∠BAC + ∠ACB = ∠CBD
80° = ∠CBD
∠CBD = 80°
Let, GH || BD
∠DEH = 35°
∠CGE = 80°
∠CGE + ∠GCE = ∠CEH
80° + 45° = ∠CEH
∠CEH = 125°
∠DEH + ∠CEH = ∠CED
125° + 35° = ∠CED
∠CED = 160°
Related Questions on Geometry
A. $$\frac{{23\sqrt {21} }}{4}$$
B. $$\frac{{15\sqrt {21} }}{4}$$
C. $$\frac{{17\sqrt {21} }}{5}$$
D. $$\frac{{23\sqrt {21} }}{5}$$
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is

A. 30°
B. 40°
C. 20°
D. 70°


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