INSCRIBED ANGLE AND INTERCEPTED ARC OF CIRCLE

Find the measure of angle indicated.

Problem 1 :

Find ∠G

Solution :

∠HGF = 1/2 arc HF

∠HGF = (1/2) 90

∠HGF = 45

Problem 2 :

Solution :

∠RTS = 1/2 arc RS

∠RTS = (1/2) 48

∠RTS = 24

Problem 3 :

Solution :

∠ABC = 1/2 arc AC

51 = (1/2) x

x = 51(2)

x = 102

Problem 4 :

Find measure of arc TV.

Solution :

∠TUV = 1/2 arc TV

38 = 1/2 arc TV

measure of arc TV = 2(38)

measure of arc TV = 76

Problem 5 :

Find measure of arc RS and m∠STR. What do you notice about m∠STR and m∠RUS.

Solution :

m∠RUS lies between chords SU and RU.

m∠RUS = 1/2 of measure of arc RS

31 = 1/2 of measure of arc RS ----(1)

m∠RTS lies between chords ST and RT.

m∠RTS = 1/2 of measure of arc RS

Applying the value of 1/2 of measure of arc RS in (1), we get

m∠RTS = 31

m∠RTS and m∠RUS are equal.

Problem 6 :

Find the following.

Solution :

∠ZYW = angle between ZY and YW

∠ZXW = angle between ZX and XW

∠ZYW = (1/2) measure of arc ZW

72 = (1/2) measure of arc ZW

Measure of arc ZW = 72(2)

Measure of arc ZW = 144

∠ZXW = (1/2) measure of arc ZW

∠ZXW = (1/2) 144

∠ZXW = 72

Problem 7 :

Find the values of x and y.

Solution :

∠ABD = ∠ACD (lies between the same arc AD)

x = 58

∠CDB = ∠BAC (lies between the same arc BC).

y = 41

Problem 8 :

Find ∠BAC

Solution :

∠BAC = (1/2) measure of arc CD

∠BAC = (1/2) (80)

∠BAC = 40

Problem 9 :

Solution :

In the triangle BDC,

∠B + ∠D + ∠C = 180

∠B + 49 + 70 = 180

∠B = 180 - 119

∠B = 61

∠DBC = (1/2) of measure of arc DC

61 = (1/2) of measure of arc DC

Measure of arc DC = 61(2)  ==> 132.

Problem 10 :

Solution :

∠RSQ = 1/2 measure of arc QR

95 = 1/2 measure of arc QR

Measures of arc QR = 95(2)

Measures of arc QR = 190

Measure of arc QR + Measure of arc RS + Measure of arc SQ = 360

190 + 75 + Measure of arc SQ = 360

Measure of arc SQ = 360 - 190 - 75

Measure of arc SQ = 95

Problem 11 :

ADB is an arc of a circle with center O, if ∠ACB = 35, find ∠AOB

Solution :

∠AOB = angle created at the center of the circle

∠ACB = angle created at the circumference of the circle.

∠AOB = 2 ∠ACB

∠AOB = 2(35)

= 70

So, the required angle measure is 70 degree.

Problem 12 :

AOB is the diameter of the circle with the center O. Is ∠APB = ∠AQB = 90 degree. Give reasons.

Solution :

Angle created at the center = ∠AOB = 180

Angle at the circumference of the circle = ∠APB

∠AOB = 2 ∠APB

180 = 2∠APB

∠APB = 180/2

= 90

Angle at the semicircle is 90 degree.

Problem 13 :

PQR is an arc of the circle with center O. If  ∠PTR = 35. find  ∠PSR

Solution :

∠PTO = 35

∠PSR = 35

Angles at the same segment.

Problem 14 :

AB and CD are two equal chords of the circle with center O. If ∠AOB = 55, find ∠COD.

Solution :

AO = OC (Radii)

OB = OD (radii)

∠AOB = ∠COD = 55

Problem 15 :

PQRD is a cyclic quadrilateral and the side S is extended to the point A. IF ∠PQR = 80, find ∠ASR.

Solution :

∠PQR = 80

∠PQR + ∠PSR = 180

80 + ∠PSR = 180

∠PSR = 180 - 80

∠PSR = 100

∠ASR = 180 - 100

= 80

So, the required angle measure ∠ASR is 80 degree.

Problem 16 :

ABCD is cyclic quadrilateral whose diagonals intersect at O. If ∠ACB = 50 and ∠ABC = 110, find ∠BDC.

Solution :

∠ADB = 50

because angle in the same segment they will be equal.

∠ADB + ∠BDC + ∠ABC = 180

50 + ∠BDC + 110 = 180

∠BDC + 160 = 180

∠BDC = 180 - 160

∠BDC = 20

Problem 17 :

AB is a chord of the circle with the center O. IF ∠ACB = 40, find ∠AOB

Solution :

∠ACB = 40

∠AOB = 2(40)

= 80

OA = OB = radii

∠OAB =  ∠OBA

∠AOB +  ∠OAB + ∠OBA = 180

80 + 2 ∠OAB = 180

2∠OAB = 180 - 80

2∠OAB = 100

∠OAB = 100/2

= 50