Question

In figure,angle ABC=120^0,where A,B and c are points on the circle with centre O.Find angle OAC​

Answer 1

∠OAC  = 30° if ∠ABC = 120° where A,B and c are points on the circle with centre O

Step-by-step explanation:

∠ABC = 120°

=> ∠AOC on outer side = 2 * 120° = 240°

=> ∠AOC in inner side = 360° - 240° = 120°

Noe letd draw OM ⊥ AC

then M will be mid point of AC

ΔOAM ≅ ΔOCM

=> ∠OAM = ∠OCM

∠AOM = ∠COM

∠AMO = ∠CMO = 90°

∠AOM +  ∠COM = ∠AOC  = 120°

=> ∠AOM = ∠COM = 60°

in Δ OAM

∠OAM + ∠AMO + ∠AOM = 180°

=> ∠OAM + 90° + 60° = 180°

=> ∠OAM = 30°

∠OAM = ∠OAC  as M lies on AC

=>  ∠OAC  = 30°

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Answer 2

The measure of ∠OAC is 30°.

Step-by-step explanation:

Given information: ∠ABC = 120°, O is the center of the circle.

According to the central angle theorem the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

Let measure of ∠OAC  is x°.

Major angle OAC = $(360-x)^{\circ}$            .... (1)

Using the central angle theorem, we get

Major angle OAC = 2\angle ABC[/tex]

Major angle OAC = 2(120)[/tex]

Major angle OAC = 240^{\cric}[/tex]                ... (2)

From (1) and (2) we get

$(360-x)=240$

$360-240=x$

$120=x$

Since OA and OC are radius of the triangle, it means OA=OC and ∠OAC=∠OCA.

The sum all interior angles or a triangle is 180 degrees.

$\angle OAC+\angle OCA+\angle AOC=180$

$\angle OAC+\angle OAC+120=180$             (∠OAC=∠OCA)

$2\angle OAC=180-120$

$2\angle OAC=60$

Divide both sides by 2.

$\angle OAC=30$

Therefore, the measure of angle OAC is 30°.

#Learn more

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