Inscribed Angle problems

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Inscribed angle

An inscribed angle has a vertex on a circle and sides (arms)  which are the chords of circle.

An intercepted arc has end points on the sides of an inscribed angle and lies in the interior of inscribed angle.

In the above figure ∠QRS is the inscribed angle and  arc QS is the intercepted arc.

Inscribed Angle theorem:

If an angle is inscribed in a circle, then the measure of angle equals one half the measure of intercepted arc.

\angle QRS = \frac{1}{2} \times m \widehat{QS}

 

Here are some problems on inscribed angle theorem:

 

In the figure below ΔABC is inscribed in the circle. Find the measure of arc AC.

Solution:

Here ∠ABC is the inscribed angle and arc AC is the intercepted arc.

That means to find arc AC, we need to find ∠ABC first.

∠ABC + 80+45 = 180          [    using triangle’s angle sum property   ]

∠ABC  = 180-125

∠ABC  = 55

\angle ABC = \frac{1}{2} \times m \widehat{AC}         [  Using inscribed angle theorem  ]

55 = \frac{1}{2} \times m \widehat{AC}

2\times 55 = m \widehat{AC}

110 = m \widehat{AC}

Therefore measure of arc AC is 110º.

 

In the figure below, points A, B, C and D are on the circle. The measure of arc AB is 65 and measure of arc BC is 70 . Find measure of angle ADC.

Solution: We know that

\angle ADB = \frac{1}{2} \times m \widehat{AB}                    [ using inscribed angle theorem  ]

\angle ADB = \frac{1}{2} \times 65 = 32.5

Similarly,

\angle BDC = \frac{1}{2} \times m \widehat{BC}

\angle BDC = \frac{1}{2} \times 70 = 35

\angle ADC= \angle ADB +\angle BDC

\angle ADC= 32.5 +35 = 67.5

 

Theorem : The angle at the centre is twice the angle at the circumference subtended by the same arc.

 

In the figure given below find angle FHG.

Solution:  We know that the angle at the centre is twice the angle at the circumference subtended by the same arc. Therefore to find angle FHG we need to first find angle at the center intercepted by arc GF.

\angle HEG +\angle HEF + \angle FEG = 360              [ complete angle around a point is 360 ]

153 + 125 + \angle FEG = 360

\angle FEG = 360-278 = 82

∠FEG at the center would be twice of ∠FHG at the circumference as they both are subtended by same arc GF.

\angle FEG = 2\angle FHG

82 = 2\angle FHG

41 = \angle FHG

\angle FHG=41^{\circ}

 

Theorem : If two inscribed angles of a circle intercept the same arc , then the angles are congruent.

 

In the following figure , find angle D and C

Solution:  Since angle D and C both are intercepted by same arc TS , so their measures would be equal.

3x-5=2x+15

3x-2x=15+5

x= 20

Angle D = 3x-5 = 3(20)-5 = 55

Angle C = 2x+15 = 2(20)+15 = 55

Therefore measure of  each angle C and D is 55.

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