Question

in the given figure BC is a chord and CD ia a tangent trough the point C. If angle AOC=118° then find angle ACD
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Answer 1

$\huge\textsf\orange{Question}$

in the given figure BC is a chord and CD ia a tangent trough the point C. If angle AOC=118° then find angle ACD

$\huge\textsf\red{ANSWER}$

$\huge\textsf\blue{Given}$

• AOC=118°

$\huge\textsf\pink{LET'S\:CONSTRUCT}$

Take any point M on the major arc of BC of a circle.

Join AM, BM, and CM

Now, we have a cyclic quadrilateral BACM

As sum of opposite angles of a quadrilateral is 180°,

∴ ∠BAC+∠BMC=180° ........ (1)

As we know, the angle subtended by a chord at the centre is double of the angle subtended by a chord on the circle

∴ ∠BOC=2∠BMC ........

(2) From (1) and (2),

∠BAC+ $\frac{1}{2}$ ∠BOC=180°.....(3)

Now, BO=OC ........ (radii of circle)

∴ ∠OBC=∠OCB

In △OBC,

∠OBC+∠OCB+∠BOC=180°

2∠OBC+∠BOC=180 °

⟹ ∠BOC=180 °

−2∠OBC ........ (4)

From (3) and (4),

∠BAC+ $\frac{1}{2}$

(180 −2∠OBC)=180°

⟹ ∠BAC+90 −∠OBC=180°

$\huge\textsf\blue{hence,}$

∴ ∠BAC−∠OBC=90°

Answer 2

Answer:

$\huge\textsf\orange{Question}$

in the given figure BC is a chord and CD ia a tangent trough the point C. If angle AOC=118° then find angle ACD

$\huge\textsf\red{ANSWER}$

$\huge\textsf\blue{Given}$

AOC=118°

$\huge\textsf\pink{LET'S\:CONSTRUCT}$

Take any point M on the major arc of BC of a circle.

Join AM, BM, and CM

Now, we have a cyclic quadrilateral BACM

As sum of opposite angles of a quadrilateral is 180°,

∴ ∠BAC+∠BMC=180° ........ (1)

As we know, the angle subtended by a chord at the centre is double of the angle subtended by a chord on the circle

∴ ∠BOC=2∠BMC ........

(2) From (1) and (2),

∠BAC+ \frac{1}{2}

2

1

∠BOC=180°.....(3)

Now, BO=OC ........ (radii of circle)

∴ ∠OBC=∠OCB

In △OBC,

∠OBC+∠OCB+∠BOC=180°

2∠OBC+∠BOC=180 °

⟹ ∠BOC=180 °

−2∠OBC ........ (4)

From (3) and (4),

∠BAC+ \frac{1}{2}

2

1

(180 −2∠OBC)=180°

⟹ ∠BAC+90 −∠OBC=180°

$\huge\textsf\blue{hence,}$

∴ ∠BAC−∠OBC=90°