(a)
In the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.
(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.
In the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.
(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option
Explanation:
(a) \(3(x + y) + 45 = 180°\) ( sum of angles on a straight line)
\(3x + 3y + 45 = 180 \implies 3x + 3y = 135\)
\(3x = 135 - 3y \implies x = 45 - y\)
Also,
\(5x + y + y = 180\)
\(5x + 2y = 180..... (*)\)
Put x = 45 - y into (*) above,
\(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\)
\(225 - 3y = 180 \implies 3y = 45\)
\(y = \frac{45}{3} = 15°\)
\(x = 45 - y\)
\(x = 45 - 15 = 30°\)
(b)
Let the distance BD = x and CD = y.
In \(\Delta ABD\),
\(\tan 53 = \frac{33}{x}\)
\(x = \frac{33}{\tan 53}\)
= \( 24.868 cm\)
In \(\Delta ACD\),
\(\tan 67 = \frac{33}{y}\)
\(y = \frac{33}{\tan 67}\)
\(y = 14.01 cm\)
\(\therefore BC = x + y \)
\(24.868 + 14.01 = 38.878 cm\)
\(\approxeq 38.9 cm\)
(a) \(3(x + y) + 45 = 180°\) ( sum of angles on a straight line)
\(3x + 3y + 45 = 180 \implies 3x + 3y = 135\)
\(3x = 135 - 3y \implies x = 45 - y\)
Also,
\(5x + y + y = 180\)
\(5x + 2y = 180..... (*)\)
Put x = 45 - y into (*) above,
\(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\)
\(225 - 3y = 180 \implies 3y = 45\)
\(y = \frac{45}{3} = 15°\)
\(x = 45 - y\)
\(x = 45 - 15 = 30°\)
(b)
Let the distance BD = x and CD = y.
In \(\Delta ABD\),
\(\tan 53 = \frac{33}{x}\)
\(x = \frac{33}{\tan 53}\)
= \( 24.868 cm\)
In \(\Delta ACD\),
\(\tan 67 = \frac{33}{y}\)
\(y = \frac{33}{\tan 67}\)
\(y = 14.01 cm\)
\(\therefore BC = x + y \)
\(24.868 + 14.01 = 38.878 cm\)
\(\approxeq 38.9 cm\)