32o
40o
60o
69o
Correct answer is D
sin 60o = x + 0.5 0o(given)
0.8660 = x + 0.5
0.8660 - 0.5 = x
x = 0.3660
cos\(\theta\) = x(given)
cos\(\theta\) = 0.3660
Hence, \(\theta\) = cos-1(0.3660)
= 68.53o
= 69o (nearest degree)
A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
\(4\sqrt{11}\)
\(4\sqrt{10}\)
\(2\sqrt{17}\)
\(2\sqrt{13}\)
Correct answer is D
|PQ| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\)
= \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\)
= \(\sqrt{4^2 + 6^2}\)
= \(\sqrt{16 + 36}\)
= \(\sqrt{52}\)
= 2\(\sqrt{13}\)
A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ
\(\frac{3}{5}\)
\(\frac{2}{3}\)
\(\frac{3}{2}\)
\(\frac{5}{3}\)
Correct answer is C
Let: (x1, y1) = (1, 2)
(x2, y2) = (5, 8)
The gradient m of \(\bar{PQ}\) is given by
m = \(\frac{y_2 y_1}{x_2 - x_1}\)
= \(\frac{8 - 2}{5 - 1}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
3
5
6
8
Correct answer is A
Volume of pyramid = \(\frac{1}{3}\)lbh
90 = \(\frac{1}{3} \times x \times 6 \times 15\)
x = \(\frac{90 \times 33}{6 \times 15}\)
= 3
2.6cm
3.5cm
3.6cm
7.0cm
Correct answer is B
Curved surface area of cylinder = 2\(\pi\)rh
110 = 2 x \(\frac{22}{7}\) x r x 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm