8.60m
7.51m
7.15m
1.19m
Correct answer is C
Tan 50o = \(\frac{h}{6}\)
h = 6 tan 50
= 6 x 1.1917
= 7.1505
= 7.15
If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x)
\(\frac{7}{13}\)
\(\frac{12}{13}\)
\(\frac{79}{156}\)
\(\frac{209}{156}\)
Correct answer is 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 = \(\frac{12}{13} - \frac{5}{12}\)
\(\frac{144 - 65}{156} = \frac{79}{156}\)
1cm
3cm
5cm
6cm
Correct answer is C
Volume of a cube with side a cm = a\(^3\)
a\(^3\) = 64 \(\implies\) a = 4cm
Base area of a cube = a\(^2\)
= 4\(^2\)
= 16 cm\(^2\)
\(\implies\) Base area of the cuboid = 16 cm\(^2\)
Volume of cuboid = Base area x height
80 = 16 x h
h = \(\frac{80}{16}\)
= 5 cm
2\(\sqrt{21}\)cm
\(\sqrt{42}\)cm
2\(\sqrt{19}\)cm
\(\sqrt{21}\)cm
Correct answer is A
From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ2
52 = 22 + MQ2
25 = 4 + MQ2
25 - 4 = MQ2
21 - MQ2
MQ2 = 21
MQ2 = \(\sqrt{21}\)
Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm
Given that p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\), make q the subject of the equation
q = p\(\sqrt{r}\)
q = p3r
q = pr3
q = pr\(\frac{1}{3}\)
Correct answer is D
p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\)(cross multiply)
3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply)
3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side
q = 3\(\sqrt{r}\) x p
q = r\(\frac{1}{3}\)p = pr\(\frac{1}{3}\)