Question

16. In Fig. 10.170, ZOAB = 30' and ZOCB = 57°. Find ZBOC and ZAOC.
17. O is the centre of the circe and chords AC and BD intersect at P such that <APE = 120 and 2PCs
Find the value of <ADB


18. In Fig. 10.172, BC is a diameter of a circle with centre O. If <BAO = 65°, find <AOC and <ADC​

Answer 1

Answer:

16)

⇒ ∠OAB=30 °and ∠OCB=57° .

In △AOB,

⇒ AO=OB [ Radius of a circle ]

⇒ AO=OB [ Radius of a circle ]⇒ ∠OBA=∠BAO=30°

[ Angles opposite to equal sides are equal ]

In △AOB,

In △AOB,⇒ ∠AOB+∠OBA+∠BAO=180°.

∴ ∠AOB+30°+30°

=180 °∴ ∠AOB=180° −30° −30 °∴ ∠AOB=120°

----- ( 1 )

Now, in △OCB,

Now, in △OCB,⇒ OC=OB [ Radius of a circle ]

Now, in △OCB,⇒ OC=OB [ Radius of a circle ]⇒ ∠OBC=∠OCB=57°. [ Angles opposite to equal sides are equal ]

[ Angles opposite to equal sides are equal ]In △OCB,

⇒ ∠BOC+∠OCB+∠OBC=180°.

⇒ ∠BOC+57°+57°=180°

⇒ ∠BOC=180°−114°

∴ ∠BOC=66°

------ ( 2 )

------ ( 2 )⇒ ∠AOB=120°

[ From ( 1 ) ]

[ From ( 1 ) ]⇒ ∠AOC+∠COB=120°

⇒ ∠AOC=120° −66°

∴ ∠AOC=54°.

18)

In △AOB

In △AOBOA=OB

=>∠OAB=∠OBA=60°

We know that exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.

We know that exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.∠AOC=∠OAB+∠OBA

120° =60°+60°

We know that angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.

We know that angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.∴∠ADC=

$\frac{1}{2}$

∠AOC

$\frac { = 120}{2}$

=60°