(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...

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).