< QPS = 30...
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
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