Question

the law of cosines: applications
there is a transmission tower near i-10. the range of the service from the tower form a 47° angle and range of service is 28 miles to one section of i-10 and 31 miles to another point on i-10. if a driver is traveling at 50 mph, how long will she have service?
a steep mountain is inclined 74 degree to the horizontal and rises to a height of 3400 ft above the surrounding plain. a cable car is to be installed running to the top of the mountain from a point 980 ft out in the plain from the base of the mountain. find the shortest length of cable needed.
your answer is ft

Explanation:

Step1: Find the base length of the mountain

Let the base length of the mountain be \(x\). Using the sine function \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), we have \(\sin(74^{\circ})=\frac{3400}{l_1}\) (where \(l_1\) is the length from the base of the mountain to the top along the slope). So \(l_1 = \frac{3400}{\sin(74^{\circ})}\approx\frac{3400}{0.9613}\approx3537\) ft. And using the cosine function \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(x = l_1\cos(74^{\circ})\approx3537\times0.2756\approx975\) ft.

Step2: Use the law of cosines

The two - side lengths of the triangle for the cable problem are \(a = 980\) ft and \(b\approx3537\) ft, and the included angle \(\theta=180^{\circ}- 74^{\circ}=106^{\circ}\). According to the law of cosines \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).
Substitute \(a = 980\), \(b\approx3537\), \(\cos\theta=\cos(106^{\circ})\approx - 0.2756\) into the formula:

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Then \(c=\sqrt{15370769}\approx3921\) ft.

Answer:

\(3921\) ft