Question

In the fig. AB and CD are two lines of center O if angle AOD =85°and angle EBE =35° Find angle ACD

Answer 1

Given:

AB and CD are two lines of center O

angle AOD = 85°and angle EBE = 35°

To find:

angle ACD

Solution:

We know that → The angle in a semicircle subtended by the diameter at the circumference is 90°.

So, here we have AB as one of the diameters of the given circle

∴ ∠AEB = 90°

In Δ ABE, by angle sum property we get

∠ABE + ∠AEB + ∠EAB = 180°

⇒ 35° + 90° + ∠EAB = 180°

⇒ ∠EAB = 180° - 125°

⇒ ∠EAB = 55°

∠AOD + ∠AOC = 180° [Linear Pairs]

⇒ 85° + ∠AOC = 180°

⇒ ∠AOC = 180° - 85°

⇒ ∠AOC = 95°

Now,

In Δ AOC, by angle sum property, we get

∠ACO + ∠AOC + ∠CAO = 180°

⇒ ∠ACO + 95° + 55° = 180° [∠CAO = ∠EAB]

⇒ ∠ACO + 150° = 180°

⇒ ∠ACO = 180° - 150°

⇒ ∠ACO = 30°

⇒ ∠ACO = ∠ACD = 30°

Thus, $\boxed{\bold{\angle ACD = \underline{30\°}}}$.

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