(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 point...
Question 1 Report
(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.
Answer Details
(a) From the diagram, AOB is a straight line, so the angles on the upper side of the line sum to \(180^\circ\), and the angles on the lower side also sum to \(180^\circ\).
Upper side (angles \(AOC\) and \(COB\)):
\[3(x+y) + 45 = 180\]\[3(x+y) = 135 \Rightarrow x + y = 45 \quad (1)\]Lower side (angles \(AOD\) and \(DOB\)):
\[(5x + y) + y = 180\]\[5x + 2y = 180 \quad (2)\]From (1), \(y = 45 - x\). Substitute into (2):
\[5x + 2(45 - x) = 180\]\[5x + 90 - 2x = 180\]\[3x = 90 \Rightarrow x = 30\]Then \(y = 45 - 30 = 15\).
\(x = 30,\quad y = 15\)
Check: \(3(x+y)=135^\circ\), \(135+45=180^\circ\) (upper). \(5x+y=165^\circ\), \(165+15=180^\circ\) (lower). Correct.
(b) Let the base of the pole be \(B\), and let the two points be \(P\) and \(Q\) on opposite sides. The pole \(BT = 33\ \text{m}\) is vertical.
For the point with elevation \(53^\circ\):
\[\tan 53^\circ = \frac{33}{PB} \Rightarrow PB = \frac{33}{\tan 53^\circ} = \frac{33}{1.3270} = 24.867\ \text{m}\]For the point with elevation \(67^\circ\):
\[\tan 67^\circ = \frac{33}{QB} \Rightarrow QB = \frac{33}{\tan 67^\circ} = \frac{33}{2.3559} = 14.008\ \text{m}\]Since the points are on opposite sides of the pole, the distance between them is:
\[PQ = PB + QB = 24.867 + 14.008 = 38.875\ \text{m}\]Distance \(\approx 38.9\ \text{m}\) (to 3 significant figures).
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