Question

1. katie has been given the task of determining the height of a building using the diagram below. she determines that ad = 23 metres, ∠bac = 48° and ∠adb = 65°. she knows that the shorter building is 78 metres tall. determine the height of the taller building to the nearest metre.

Explanation:

Step1: Find the length of \( BD \)

In right - triangle \( ABD \), we know \( AD = 23\) m and \(\angle ADB=65^{\circ}\). Using the trigonometric ratio \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), we have \(\tan65^{\circ}=\frac{AB}{BD}\). But also, in right - triangle \( ABD \), using the Pythagorean theorem is not the best approach. Instead, using the sine and cosine relations. Wait, better use \(\tan\angle ADB=\frac{AB}{BD}\), no. Wait, we can use the fact that in \(\triangle ABD\), \(\sin\angle ADB=\frac{AB}{AD}\) (no, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)). Wait, correct: In right - triangle \( ABD\), \(\sin65^{\circ}=\frac{AB}{AD}\), so \(AB = AD\sin65^{\circ}\approx23\times0.9063 = 20.8449\) m. And \(\cos65^{\circ}=\frac{BD}{AD}\), so \(BD=AD\cos65^{\circ}\approx23\times0.4226 = 9.72\) m.

Step2: Find the length of \( BC \)

In right - triangle \( ABC\), \(\tan\angle BAC=\frac{BC}{AB}\). Given \(\angle BAC = 48^{\circ}\) and \(AB\approx20.8449\) m. Then \(BC = AB\tan48^{\circ}\). Since \(\tan48^{\circ}\approx1.1106\), \(BC\approx20.8449\times1.1106\approx23.15\) m.

Step3: Calculate the height of the taller building

The height of the taller building \(h=BC + 78\). Substitute \(BC\approx23.15\) m. Then \(h\approx23.15+78=101.15\approx101\) m.

Answer:

\(101\) m