i the diagram given PQ 10cm PS 8cm and lt PSR i... - JAMB Mathematics 2006 Question
i the diagram given PQ = 10cm, PS = 8cm and < PSR is 60 while < SRQ is a right angle. Find SR
A
10 \(\sqrt{3}\)cm
B
10cm
C
45cm
D
14cm
correct option: d
Draw a line perpendicular to |SR| to form |PT| in \(\bigtriangleup\) PST, cos 60 = \(\frac{|ST|}{8}\(
|PT| = 8 cos 6 = 4cm
Since |TR| = |PQ|
SR = ST + TR
= (4 + 10)cm
= 14cm
|PT| = 8 cos 6 = 4cm
Since |TR| = |PQ|
SR = ST + TR
= (4 + 10)cm
= 14cm
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