Question

mountain officials want to build a new ski lift from b to c, as shown in the figure below. the distance from a to c is 1480 feet. they measure angle dac to be 36° and angle dbc to be 25°. what is the distance from a to b? round your answer to the nearest tenth of a foot.

Explanation:

Step1: Find the length of \(BC\)

In \(\triangle ADC\), \(\angle DAC = 36^{\circ}\), \(AC = 1480\) feet.
Using the sine - rule in \(\triangle BDC\) (noting that \(\angle BCA=\angle DAC-\angle DBC = 36^{\circ}- 25^{\circ}=11^{\circ}\))
By the sine - rule \(\frac{BC}{\sin\angle DAC}=\frac{AC}{\sin\angle DBC}\)
\(BC=\frac{AC\times\sin\angle DAC}{\sin\angle DBC}\)
Substitute \(AC = 1480\), \(\angle DAC = 36^{\circ}\), \(\angle DBC=25^{\circ}\)
\(BC=\frac{1480\times\sin36^{\circ}}{\sin25^{\circ}}\)
Since \(\sin36^{\circ}\approx0.5878\), \(\sin25^{\circ}\approx0.4226\)
\(BC=\frac{1480\times0.5878}{0.4226}=\frac{870.944}{0.4226}\approx2060.9\)

Step2: Find the length of \(AB\)

Again, using the sine - rule in \(\triangle ABC\), \(\angle BCA = 11^{\circ}\), \(\angle DBC = 25^{\circ}\)
By the sine - rule \(\frac{AB}{\sin\angle BCA}=\frac{BC}{\sin\angle BAC}\) (where \(\angle BAC = 180^{\circ}-36^{\circ}=144^{\circ}\), \(\angle BCA = 11^{\circ}\), \(BC\approx2060.9\))
\(AB=\frac{BC\times\sin\angle BCA}{\sin(180^{\circ}-\angle DAC)}\)
Since \(\sin(180^{\circ}-\alpha)=\sin\alpha\) (here \(\alpha = 36^{\circ}\)), \(\sin11^{\circ}\approx0.1908\), \(\sin36^{\circ}\approx0.5878\)
\(AB=\frac{2060.9\times0.1908}{0.5878}=\frac{393.22}{0.5878}\approx669.0\)

Answer:

\(669.0\)