What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?
\(\frac{\sqrt{5}}{2}\)
\(\frac{\sqrt{5}}{20}\)
\(\frac{5}{\sqrt{13}}\)
6
Correct answer is A
2x - 4y + 3 = 0
Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)
= \(\frac{4 - 12 + 3}{\sqrt{20}}\)
= \(\frac{-5}{-2\sqrt{5}}\)
= \(\frac{\sqrt{5}}{2}\)
240o
210o
150o
60o
Correct answer is B
Using sinØ = \(\frac{6√3}{12}\) → \(\frac{√3}{2}\)
sinØ = \(\frac{√3}{2}\) or 60°
The bearing of Z from X = [270 - 60]° → 210°
If in the diagram, FG is parallel to KM, find the value of x
75o
95o
105o
125o
Correct answer is B
No explanation has been provided for this answer.
\(\frac{800\pi}{9}\)km
\(\frac{800\sqrt{3\pi}}{9}\)km
800\(\pi\) km
800\(\sqrt{3\pi}\) km
Correct answer is C
Angular difference (\(\theta\))= 25° + 20° = 45°
\(\alpha\) = common latitude = 60°
\(S = \frac{\theta}{360°} \times 2\pi R \cos \alpha\)
\(S = \frac{45°}{360°} \times 2 \pi \times 6400 \times \cos 60°\)
= \(\frac{6400\pi}{8} = 800\pi km\)
If the angles of quadrilateral are (p + 10)°, (2p - 30)°, (3p + 20)° and 4p°, find p.
63
40
36
28
Correct answer is C
The sum of angles in a quadrilateral = 360°
\(\therefore (p + 10) + (2p - 30) + (3p + 20) + 4p = 360\)
\(10p = 360° \implies p = \frac{360}{10} = 36°\)
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