Question

3. In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that
AED = 138° and BAE = 75°. Find ECD. [1]

Answer 1

Answer:

∠BEC+∠DEC=180  

∘

 (complimentary angles)  

130  

∘

+∠DEC=180  

∘

 

∠DEC=50  

∘

 

In ΔDEC -  

∠DEC+∠ECD+∠CDE=180  

∘

 

∠CDE=110  

∘

 

by theorem of circles -

∠BAC=∠CDE

∴∠BAC=110  

∘

 

Step-by-step explanation: