The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∠POQ = 180
2∠PAQ = ∠POQ
2∠PAQ = 180
∠PAQ = 180/2
∠PAQ = 90
Problem 1 :
P, Q and R are points on the circumference of a circle, center O. PR is the diameter of the circle and ST is a tangent to the circle at the point R.
∠QRS = 58
a) Work out the size of angle QRP.
b) Work out the size of the angle QPR.
Solution :
∠ORS = 90
a) Work out the size of angle QRP :
∠ORQ + ∠ORS = 90
∠ORQ + 58 = 90
∠ORQ = 90 - 58
∠ORQ = 32
∠QRP = 32
b) Work out the size of the ∠QPR :
∠QPR = 180 - (∠QRP + ∠PQR)
∠QPR = 180 - (32 + 90)
∠QPR = 180 - 122
∠QPR = 58
Using alternate segment theorem, we can prove ∠QPR = 58.
Problem 2 :
Work out the size of each angle marked with a letter. Give reasons for your answers.
Solution :
∠OAP = 90
∠OAB + ∠BAP = 90 ---(1)
In triangle OAB,
∠OBA + ∠OAB + ∠AOB = 180
Let ∠OBA = x
x + 132 + x = 180
2x + 132 = 180
2x = 180 - 132
2x = 48
x = 48/2
x = 24
By applying x = 24 in (1), we get
24 + ∠BAP = 90
∠BAP = 90 - 24
b = ∠BAP = 66
Problem 3 :
In the circle given below, U is the point of ST such that RUT is right angle.
find
a) ∠URT b) ∠ROT c) ∠ORU d) ∠ORT
Solution :
In the figure above,
∠SRT = 90 (angle in a semicircle)
In triangle RST,
∠RST + ∠SRT + ∠STR = 180
37 + 90 + ∠STR = 180
127 + ∠STR = 180
∠STR = 180 - 127
∠STR = 53
a) Finding ∠URT :
In triangle URT,
∠RUT + ∠TRU + ∠UTR = 180
90 + ∠TRU + 53 = 180
143 + ∠TRU = 180
∠TRU = 180 - 143
∠TRU = 37
b) Finding ∠ROT :
Since ∠OSR = ∠ORS = 37
∠ROT = 37 + 37 ==> 74
c) Finding ∠ORU :
∠SRO + ∠ORU + ∠URT = 90
37 + ∠ORU + 37 = 90
74 + ∠ORU = 90
∠ORU = 90 - 74
∠ORU = 16
d) Finding ∠ORT :
∠ORT = ∠ORU + ∠URT
∠ORT = 16 + 37
∠ORT = 53
Problem 4 :
Solution :
In the triangle inside the circle, the angle in the semicircle is right angle.
55 + 90 + g(alternate segment) = 180
145 + g = 180
g = 180 - 145
g = 35
Problem 5 :
Work out the size of each angle marked with a letter. Give reasons for your answers.
Solution :
∠PBA = ∠PAB = g
Because the tangent drawn from the external point of the circle will be equal in measure. So, its length will also will have same angle measure.
∠BPA + ∠PBA + ∠PAB = 180
56 + g + g = 180
56 + 2g = 180
2g = 180 - 56
2g = 124
g = 62
Since OB is the line drawn from center of the circle to the point of contact of the tangent.
g + h = 90
62 + h = 90
h = 90 - 62
h = 28
Problem 6 :
In the given figure O is the center of the circle and SP is a tangent. If ∠SRT = 65, find the values of x, y and and z.
Solution :
In triangle STR,
∠SRT + ∠STR + ∠TSR = 180
65 + x + 90 = 180
x + 155 = 180
x = 180 - 155
x = 25
y = 2(25) ==> 50
In triangle SOP,
∠OSP + ∠SPO + ∠POS = 180
90 + z + y = 180
90 + z + 50 = 180
140 + z = 180
z = 180 - 140
z = 40
Problem 7 :
Work out the size of each angle marked with a letter. Give reasons for your answers.
Solution :
From the picture, it is clear d = 36.
In the triangle,
e + d + 36 = 180
e + 36 + 36 = 180
e = 180 - 72
e = 108
f + d = 90
f + 36 = 90
f = 90 - 36
f = 54
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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