Question

) In the given figure, O is the centre of the circle,
angle BAD = 75° and chord BC = chord CD.
Find (i) angle BOC (ii) angle OBD, (iii) angle BCD.
​

Answer 1

i)∠BOC =45°,ii)∠OBD=67.5°,iii)∠BCD=75°

Step-by-step explanation:

Given,∠BAD=75°,BC=CD

Construction:Join BD and OC.

∠BOD=2∠BAD [∵Angle subtended at centre is double the angle subtended on circumference]

∠BOD=2×75°

∠BOD=150°

Similarly,∠BOD=2∠BCD

               ∠BCD=1/2∠BOD

                         =1/2×150°

              ∠BCD=75°...............(1)

Now,In ΔBOD,

OB=OD[Radius of circle ]

⇒∠OBD=∠ODB...........(2) [Angle opposite to equal side are always equal]

Now,In ΔOBD,

∠OBD+∠ODB+∠BOD=180° {By angle sum property of triangle]

∠ODB+∠ODB+150°=180° [From equation(2)]

2∠ODB=180°-150°

2∠ODB=30°

∠ODB=30°/2

∠ODB=15°=∠OBD

Now,it is given that,BC=CD

⇒∠CBD=∠CDB [Angle opposite to equal side are always equal]

Also,In ΔBCD,

∠BCD+∠CDB+∠CBD=180°[By angle sum property of triangle]

∠BCD+∠CDB+∠CDB=180°

75°+2∠CDB=180°

2∠CDB=180°-75°

∠CDB=$\frac{105}{2}$°

        52.5°

∠CDB=∠CBD=52.5°

Now,∠OBD=∠OBD+∠CBD

                 =15°+52.5°

        ∠OBD=67.5°

Also,∠OBD=∠OCD [OB=OC,radius]

In ΔOBC,

∠OBC+∠OCB+∠BOC=180°[By angle sum property]

67.5°+67.5°+∠BOC=180°

135°+∠BOC=180°

∠BOC=180°-135°

∠BOC =45°

Answer 2

The value of $\angle BOC,\angle OBD\ and\ \angle BCD$ are $75\ degree,15\ degree\ and\ \angle 105\ degree$

Step-by-step explanation:

Given,

$O$ is the center of the circle, $\angle BAD=75\ degree$ and cord$BC=$cord$CD$

From figure,

(i): $\angle BOD=2\angle BAD$

⇒$\angle BOD=2\times 75=150\ degree$

∵ cord$BC=$cord$CD$

∴$\angle BOC=\angle COD$

∴$\angle BOD=2\angle BOC$

⇒$\angle BOC=\frac{1}{2}\times \angle BOD$

⇒$\angle BOC=\frac{1}{2}\times 150$

⇒$\angle BOC=\frac{150}{2}$

⇒$\angle BOC=75\ degree$

(ii): $\angle OBD =\frac{1}{2} (180-\angle BOD)$

⇒$\angle OBD =\frac{1}{2} (180-150)$

⇒$\angle OBD =\frac{1}{2} \times 30$

∴ $\angle OBD=15\ degree$

(iii): $\angle BCD = 180-\angle BAD$

⇒$\angle BCD = 180-75$

⇒$\angle BCD = 105\ degree$

So, The value of $\angle BOC,\angle OBD\ and\ \angle BCD$ are $75\ degree,15\ degree\ and\ \angle 105\ degree$