Find the size of the angle marked x in the diagram.
108o
112o
128o
142o
Correct answer is C
x + 52o = 90
x = 90 - 52
x = 38o
k = opposite angle Z
k = 38o
y + k = 90o
y + 38o = 90o
y = 90o - 38o
y = 52o
y = x = 180o(sum of angles on straight line)
52 + x = 180o
x = 180 - 52
x = 128
In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x
60o
55o
50o
45o
Correct answer is A
< Q = \(\frac{240}{2}\) (angle at centre twice that at the circumference)
< Q = 120o
Also < POR = 360 - 240
= 120o
( < s at centre) since /PQ/ = /QR/, < x = < R
Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)
x + R + 120 = 360o
x + R = 360 - 240 = 120; Since x = R
x + x = 120
2x = 120
Since x = R
x + x = 120
2x = 120
x = \(\frac{120}{2}\)
= 60o
\(\bigtriangleup\) PQK
\(\bigtriangleup\) PKS
\(\bigtriangleup\) SKR
\(\bigtriangleup\) PSR
Correct answer is C
No explanation has been provided for this answer.
9o
10o
11o
12o
Correct answer is B
tan \(\theta\) = \(\frac{9}{50}\) = 0.18
\(\theta = tan^{-1} 0.18\)
\(\theta\) = 10.20
If 27x = 9y. Find the value of \(\frac{x}{y}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
1\(\frac{1}{2}\)
3
Correct answer is B
27x = 9y
33x = 32y
3x = 2y
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