Question

AB||CO AOD is straight line if ∆ADB=40° Find ∆BCO​

Answer 1

Answer:

Given−

ThediameterCD&anoterextendeddiameterAB

intersectatthecentreO.

TwolinesDB&ABintersectatB.

ThechordCEmeetsABatE.

∠AOD=75

o

and∠OCE=40

o

.

Tofindout−

(i)∠CDE=?

(ii)∠OBE=?

Solution−

DCisadiameter.

So∠DEC=Theangleinasemicircle=90

o

.

∴∠CDE=180

o

−(∠OCE+∠DEC)

(anglesumpropertyoftriangles)

=180

o

−(40

o

+90

o

)=50

o

.

Again∠AOD=∠COB=75

o

and∠AOC=∠DOB

(bothpairsareverticallyoppositeangles).

i.e∠AOC+∠DOB=2∠DOB

So∠AOD+∠COB+(∠AOC+∠DOB)=360

o

(completeangle)

⟹2∠DOB=360

o

−(∠AOD+∠COB)=360

o

−(75

o

+75

o

)=210

o

⟹∠DOB=105

o

.

NowinΔODB

∠OBE=180

o

−(∠DOB+∠CDE)(anglesumpropertyoftriangles)

⟹∠OBE=180

o

−(105

o

+50

o

)=25

o

.

So∠CDEand∠OBE are respectively

50

o

and25

o

.

Step-by-step explanation:

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