Question

In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O; and ∠DAC = 40° , ∠CAB = 35°; and ∠DOC = 110°. Calculate the measure of ∠ABO, ∠ADC, ∠ACB, and ∠CBD.

Answer 1

Given, ABCD is a parallelogram having ∠BAO = 35°, ∠DAO = 40° and ∠COD = 105°

Now, ∠COD = ∠AOB =  105° [vertically opposite angles]

In ΔAOB, by angle sum property of triangle, 

⇒ ∠AOB + ∠OAB + ∠ABO =  180°

⇒ 105° + 35° + ∠ABO =  180°

⇒ ∠ABO =  40°

Again, adjacent angles of a parallelogram are supplementary.

⇒ ∠DAB + ∠ABC =  180°

⇒ ∠DAO + ∠OAB + ∠ABO + ∠CBO  =  180°

⇒ 40° + 35° + 40° + ∠CBO  =  180°

⇒ ∠CBO  = ∠CBD =  180° - 115° = 65° 

⇒ ∠CBD =  65° 

In ΔABC, by angle sum property of triangle, 

⇒ ∠CAB + ∠ABC + ∠ACB =  180°

⇒ 35° + ∠ABO + ∠CBO + ∠ACB =  180°

⇒ 35° + 40° + 65° + ∠ACB =  180°

⇒ ∠ACB = 180° - 140° = 40° 

⇒ ∠ACB = 40° 

Now, opposite angles of a parallelogram are equal

⇒ ∠A =∠C 

⇒ ∠C  =  75°

On applying angle sum property of triangle in BCD, we get

⇒ ∠C  + ∠CBD + ∠CDB =  180°

⇒ 75°  + 65° + ∠CDB =  180°

⇒ ∠CDB =  40°

or ∠ODC =  40°

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Answer 2

<ACB=<DAC=40°[alt.int. angle]
<ADC+<DAB=180°[CO- int. angle]
<ADC+75°=180°
<ADC=105°
<AOB=<DOC=110°[V.O.A]
<ABO+<AOB+<OAB=180°(A.S.P)
<ABO+110°+35°=180°
<ABO=180°-145°
<ABO=35°
<CBD=40°[as <ABO=<OAB
THEREFORE<DAC=<CBD]
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