If sin x frac 5 13 and 0o leq x leq 90o find th... - WAEC Mathematics 2013 Question
If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x)
A
\(\frac{7}{13}\)
B
\(\frac{12}{13}\)
C
\(\frac{79}{156}\)
D
\(\frac{209}{156}\)
correct option: c
Sin x = \(\frac{5}{13}\)
0o \(\leq\) x \(\leq\) 90o, (cos x - tan x)
AC2 = AB2 + BC2
132 = 52 + BC2
169 - 25 + BC2
169 - 25 = BC2
144 = BC2
Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)
BC = \(\sqrt{144}\)
BC = 12
tan x = \(\frac{opp}{adj} = \frac{5}{12}\)
BC = 12
cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)
\(\frac{144 - 65}{156} = \frac{79}{156}\)
0o \(\leq\) x \(\leq\) 90o, (cos x - tan x)
AC2 = AB2 + BC2
132 = 52 + BC2
169 - 25 + BC2
169 - 25 = BC2
144 = BC2
Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)
BC = \(\sqrt{144}\)
BC = 12
tan x = \(\frac{opp}{adj} = \frac{5}{12}\)
BC = 12
cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)
\(\frac{144 - 65}{156} = \frac{79}{156}\)
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