Angles of elevation and depression Problem set B Worked solutions

A radio mast is seen from a point 26 m from its foot. The angle of elevation of the top of the mast is 36°, and the observer's eyes are 1.6 m above the ground. Find the height of the mast.

• \(\tan 36^\circ = \dfrac{h}{26}\), \(h = 18.89\)
• Add the eye height: 18.89 + 1.6 = 20.5 m

From a bridge 30 m above a river, the angle of depression of a canoe is 17°. Find the horizontal distance of the canoe from the point on the water below the bridge.

• The angle of depression from the top equals the angle of elevation from the boat, so the angle at the boat is 17°.
• \(\tan 17^\circ = \dfrac{30}{d}\), \(d = 98.13\) m

From a point A the angle of elevation of the top of a chimney is 36°. From a point 30 m further from the chimney, on the same level and in line with it, the angle of elevation is 22°. The observer's eyes are 1.6 m above the ground. Find the height of the chimney.

• \(\angle ABT = 22^\circ\), \(\angle BAT = 180^\circ - 36^\circ = 144^\circ\), \(\angle ATB = 36^\circ - 22^\circ = 14^\circ\)
• \(\dfrac{AT}{\sin 22^\circ} = \dfrac{30}{\sin 14^\circ}\), so \(AT = 46.45\)
• \(h = AT \sin 36^\circ = 46.45 \times \sin 36^\circ = 27.30\)
• Add the eye height: 27.30 + 1.6 = 28.9

The angle of elevation of the top of a wind turbine from a point A is 44°. Mia cycles 80 m directly away from the turbine to a point B, where the angle of elevation is 29°. Her eyes are 1.4 m above the ground. Find the height of the top of the turbine.

• \(\angle ABT = 29^\circ\), \(\angle BAT = 180^\circ - 44^\circ = 136^\circ\), \(\angle ATB = 44^\circ - 29^\circ = 15^\circ\)
• \(\dfrac{AT}{\sin 29^\circ} = \dfrac{80}{\sin 15^\circ}\), so \(AT = 149.9\)
• \(h = AT \sin 44^\circ = 149.9 \times \sin 44^\circ = 104.1\)
• Add the eye height: 104.1 + 1.4 = 105

From a lookout tower the angle of depression of a ship is 18°. The ship sails 250 m directly towards the tower and the angle of depression is then 33°. Find the height of the lookout above the sea.

• \(\angle ABT = 18^\circ\), \(\angle BAT = 180^\circ - 33^\circ = 147^\circ\), \(\angle ATB = 33^\circ - 18^\circ = 15^\circ\)
• \(\dfrac{AT}{\sin 18^\circ} = \dfrac{250}{\sin 15^\circ}\), so \(AT = 298.5\)
• \(h = AT \sin 33^\circ = 298.5 \times \sin 33^\circ = 162.6\)

The top of a pylon is 18 m above an observer's eye level. The angle of elevation of the top is 35° from one point. The observer walks directly away from the pylon, and the angle of elevation of the top is then 20°. How far does the observer walk?

• \(\text{First distance} = \dfrac{18}{\tan 35^\circ} = 25.71\)
• \(\text{Second distance} = \dfrac{18}{\tan 20^\circ} = 49.45\)
• Distance walked: \(49.45 - 25.71 = 23.7\) m