# Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 88: 34

The height of the top of Mount Whitney above the level road is 2.5 km

#### Work Step by Step

Let $x$ be the distance from the vertical line to the point with an elevation of $22.667^{\circ}$. We can write an expression for the height $h$: $\frac{h}{x} = tan~22.667^{\circ}$ $h = x~tan~22.667^{\circ}$ We can use the second point to write another equation for the height $h$: $\frac{h}{x+7} = tan~10.833^{\circ}$ $h = (x+7)~tan~10.833^{\circ}$ We can equate the two expressions to find $x$: $x~tan~22.667^{\circ} = (x+7)~(tan~10.833^{\circ})$ $0.4176~x = 0.1914~x+1.3395$ $0.4176~x - 0.1914~x = 1.3395$ $x = \frac{1.3395}{0.2262}$ $x = 5.9218~km$ We can use the first equation to find $h$: $h = x~tan~22.667^{\circ}$ $h = (5.9218~km)~tan~22.667^{\circ}$ $h = 2.5~km$ The height of the top of Mount Whitney above the level road is 2.5 km

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