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