Question Number 115166 by I want to learn more last updated on 24/Sep/20

Answered by 1549442205PVT last updated on 24/Sep/20

77°

Commented byI want to learn more last updated on 24/Sep/20

please workings sir

Answered by 1549442205PVT last updated on 24/Sep/20

Commented by1549442205PVT last updated on 24/Sep/20

Draw the circum circle (O)of ΔABC  and denote K=(O)∩(DE),L=AB∩CK  H=(O)∩(CE)  From the hypothesis ΔCDE is right  at E and DCE^(�) =26°⇒CDK^(�) =64°,  ΔACB is right at C and ABC^(�) =58°  ⇒BAC^(�) =32°=(1/2)CDK^(�) ⇒CBK^(⌢) =2BC^(⌢)   ⇒BC^(⌢) =BK^(⌢)  ⇒AB⊥CK at L(the property  of the diameter of a circle)and BC=BK,LC=LK  It follows that the triangles BCK and  ECK are isosceles at B and E.Hence,  AED^(�) =AEH^(�) =BEC^(�) =BEK^(�) =ϕ  ⇒AD^(⌢) =AH^(⌢) ⇒CA is the bisector of  the angle  DCE^(�) =26°⇒ACD^(�) =ACH^(�) =13°  ⇒α=ACB^(�) −ACH^(�) =90°−13°=77°  Answer α=77°

Commented byI want to learn more last updated on 24/Sep/20

Thanks sir