Question

In the given figure , O is the centre of the circle, angle ACO=35º and angle ABO = 45°, then calculate the
value of angle BOC.
​

Answer 1

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

∠BAD=90

o

Consider △BAD

Using the angle sum property

∠ADB+∠BAD+∠ABD=180

o

By substituting the values

∠ADB+90

o

+35

o

=180

o

On further calculation

∠ADB=180

o

−90

o

−35

o

By subtraction

∠ADB=180

o

−125

o

So we get

∠ADB=55

o

We know that the angle in the same segment of a circle are equal

∠ACB=∠ADB=55

o

So we get

∠ACB=55

o

Therefore, ∠ACB=55

o

Answer 2

$\underline{\underline{\huge{\gray{\tt{\textbf Given :-}}}}}$

• O is the center of a circle
• < ACO=35 and < ABO=45

$\underline{\underline{\huge{\red{\tt{\textbf Find :-}}}}}$

• angle BOC=??

$\underline{\underline{\huge{\blue{\tt{\textbf Solution :-}}}}}$

From the given information, we have the data as follows.

• In the figure, O is the centre of a circle, angle ACO = 35° and angle ABO = 45°

Here we will use the property of triangles.

• The sum of the interior angles is equal to the exterior angle.

So, in our case, we have,

∠ ACO + ∠ ABO = ∠ BOC

35° + 45° = 80°

• Therefore, the value of the angle BOC is 80°

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