Question

example 1 two students want to determine the heights of two buildings. they stand on the roof of the shorter building. the students use a clinometer to measure the angle of elevation of the top of the taller building. the angle is 44°, from the same position, the students measure the angle of depression of the base of the taller building. the angle is 53°, the students then measure the horizontal distance between the two buildings. the distance is 18.0 m. the students drew this diagram. how tall is each building? example 2 how would you calculate the length of ad using the information provided? show all your steps.

Explanation:

Step1: Find height of shorter building

In the right - triangle formed by the shorter building, the horizontal distance and the line of sight to the base of the taller building. Let the height of the shorter building be \(h_1\). We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). The angle of depression to the base of the taller building from the top of the shorter building is \(53^{\circ}\) and the horizontal distance \(BD = 18.0\)m. Since the angle of depression is equal to the angle of elevation from the base of the taller building to the top of the shorter building. So \(h_1=18\times\tan53^{\circ}\approx18\times1.3270 = 23.886\)m.

Step2: Find height of the part of taller building above the shorter building

Let the height of the part of the taller building above the shorter building be \(h_2\). The angle of elevation of the top of the taller building from the top of the shorter building is \(44^{\circ}\) and the horizontal distance \(BD = 18.0\)m. Using the tangent function, \(h_2 = 18\times\tan44^{\circ}\approx18\times0.9657=17.3826\)m.

Step3: Find height of taller building

The height of the taller building \(H=h_1 + h_2\). \(H=18\times\tan53^{\circ}+18\times\tan44^{\circ}=18(\tan53^{\circ}+\tan44^{\circ})\approx23.886 + 17.3826=41.2686\)m.

For Example 2:

Step1: Consider right - triangle \(ABC\)

In right - triangle \(ABC\), \(\tan\angle ACB=\frac{AB}{BC}\). We know \(BC = 120\)m and \(\angle ACB = 20^{\circ}\), so \(AB = 120\times\tan20^{\circ}\approx120\times0.3640 = 43.68\)m.

Step2: Consider right - triangle \(DBC\)

In right - triangle \(DBC\), \(\tan\angle DCB=\frac{DB}{BC}\). We know \(BC = 120\)m and \(\angle DCB=30^{\circ}\), so \(DB = 120\times\tan30^{\circ}=120\times\frac{\sqrt{3}}{3}=40\sqrt{3}\approx69.28\)m.

Step3: Find length of \(AD\)

\(AD=DB - AB\). \(AD = 120\times\tan30^{\circ}-120\times\tan20^{\circ}=120(\tan30^{\circ}-\tan20^{\circ})\approx69.28 - 43.68=25.6\)m.

Answer:

Height of shorter building in Example 1: approximately \(23.89\)m.
Height of taller building in Example 1: approximately \(41.27\)m.
Length of \(AD\) in Example 2: approximately \(25.6\)m.