Using the diagram, find the bearing of X from Y
300o
240o
120o
60o
Correct answer is B
The bearing of X from Y
= 360 - 120 = 240
In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x
5.6
6.5
6.6
6.8
Correct answer is A
\(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\)
\(\frac{x}{4 + x} = \frac{7}{7 + 5}\)
\(\frac{x}{4 + x} = \frac{7}{12}\)
(2x = 7)(4 + x); 12x = 28 + 7x
12x - 7x = 28
5x = 28
x = \(\frac{28}{5}\)
x = 5.6
Find the value of x in the diagram
50o
30o
22o
17o
Correct answer is D
Sum of exterior angles of a polygon = 360o
(x + 64) + 2x + (3x - 54) + 5x + 4x + 3x + (2x + 10) = 360o
20x + 20 = 360o
20x + 360 - 20 = 340o
x = \(\frac{340}{20}\)
x = 17o
In the diagram, IG is parallel to JE, JEF = 120o and FHG = 130o, fins the angle marked t
40o
70o
80o
100o
Correct answer is B
Now, t + 60o + 50o = 180o
t = 180o - 110o
t = 70o
In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP
120o
90o
60o
45o
Correct answer is B
Tan \(\theta = \frac{6}{6} = 1\)
\(\theta\) = tab - 1(1) = 45o
< top = 20
= 2 x 45o = 90o
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