If angle \(\theta\) is 135°, evaluate cos\(\theta\)
\(\frac{1}{2}\)
\(\frac{\sqrt{2}}{2}\)
\(-\frac{\sqrt{2}}{2}\)
\(-\frac{1}{2}\)
Correct answer is C
\(\theta\) = 135°
Cos 135° = Cos(90 + 45)°
= cos90° cos45° - sin90° sin45°
= 0cos45° - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
Find the equation of the line through the points (-2, 1) and (-\(\frac{1}{2}\), 4)
y = 2x - 3
y = 2x + 5
y = 3x - 2
y = 2x + 1
Correct answer is B
\(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{y - 1}{x - -2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is
\(\sqrt{13}\)
\(3\sqrt{2}\)
\(\sqrt{26}\)
\(10\sqrt{5}\)
Correct answer is B
P1 (4, 3), P2 (x, y)
y = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)
The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is
\(\frac{1}{2}\)
\(\frac{2}{5}\)
\(-\frac{4}{7}\)
\(-\frac{2}{3}\)
Correct answer is C
PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
3.92cm3
2.13cm3
1.97cm3
1.62cm3
Correct answer is D
Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3