(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he ...

(a) Height risen up the slope

The 100 m is the distance along the slope (the hypotenuse), and the height risen is the vertical (opposite) side of a right triangle whose angle to the horizontal is \(24^\circ\).

Let \(h\) be the height.

\[\sin 24^\circ = \frac{h}{100}\]

\[h = 100 \times \sin 24^\circ = 100 \times 0.4067 = 40.67\text{ m}\]

Correct to the nearest metre, \(h \approx \mathbf{41\text{ m}}\).

(b) Angle BAC of the isosceles triangle

From the diagram, \(|AB| = |AC| = 5\text{ cm}\) and \(|BC| = 8\text{ cm}\). Drop a perpendicular from \(A\) to the midpoint \(M\) of \(BC\). Then \(|BM| = \tfrac{1}{2}\times 8 = 4\text{ cm}\), and \(AM\) bisects \(\angle BAC\).

In right triangle \(ABM\):

\[\sin(\angle BAM) = \frac{BM}{AB} = \frac{4}{5} = 0.8\]

\[\angle BAM = \sin^{-1}(0.8) = 53.13^\circ\]

Therefore

\[\angle BAC = 2 \times 53.13^\circ = 106.26^\circ \approx \mathbf{106^\circ}\]

(c) Height of the light-house

The two boats are on opposite sides of the light-house and in a straight line with its foot, 70 m apart. Let the foot of the light-house be \(F\), the height be \(h\), the horizontal distance to the boat with elevation \(71.6^\circ\) be \(d_1\), and to the boat with elevation \(45^\circ\) be \(d_2\).

\[d_1 + d_2 = 70\]

From the \(71.6^\circ\) boat: \(\tan 71.6^\circ = \dfrac{h}{d_1} = 3\), so \(d_1 = \dfrac{h}{3}\).

From the \(45^\circ\) boat: \(\tan 45^\circ = \dfrac{h}{d_2} = 1\), so \(d_2 = h\).

Substitute:

\[\frac{h}{3} + h = 70\]

\[\frac{h + 3h}{3} = 70 \quad\Rightarrow\quad \frac{4h}{3} = 70\]

\[h = \frac{70 \times 3}{4} = \frac{210}{4} = 52.5\text{ m}\]

The height of the light-house is \(\mathbf{52.5\text{ m}}\).