i and iv
ii
iii
iv
Correct answer is C
m + y + x = 180o(sum of < s on straight line)
y = n(vertically opp. angle)
m + n + x = 180o
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
150o
110o
100o
95o
Correct answer is C
L + 30o - 180o(Sum of < s on straight line)
L = 180o - 30o = 150o
L = q = 150o(opposite < s are equal)
y = b = 30o(alt. < s)
b + c = 180o(sum of < s on str. line)
30o + c 180
c = 180 - 30
c = 150o
b = a = 30o (opp < s are equal)
c = d = 150o (opp < s are equal)
a + k + 70o = 180o (sum of < s on \(\bigtriangleup\))
30o + k + 70o = 180
k + 100o = 180
k = 180 - 100
k = 80o
x + 80o = 180(sum of < s on straight line)
x = 180o - 80o
x = 100o
The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n
m = \(\frac{n}{2}\)
m = 2n
m = n - 2
m = n + 2
Correct answer is A
If < SPR = 2m then < SQR = 2m but < AQR was n
n = 2m
m = \(\frac{n}{2}\)
In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x
108o
172o
130o
50o
Correct answer is B
No explanation has been provided for this answer.
In the diagram, |QR| = 10m, |SR| = 8m
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
1.32m
6.32m
9.32m
17.32
Correct answer is C
In right angled \(\bigtriangleup\)QPR
tan 30o = \(\frac{10}{x + 8}\)
(x + 8) tan 30 = 10
x + 8 = \(\frac{10}{0.5773}\)
x +8 = 17.3
x = 17.3 - 8
x = 9.32