Question

the diagram shows the portion of earth you can see when you are on top of mount rainier at point b on a clear day, where mcd ≈ 4°. what is your height above sea level? round to the nearest hundredth. 4000 mi about mi

Explanation:

Step1: Find the measure of \(\angle BAD\)

Since \(m\widehat{CD} = 4^{\circ}\), and the central angle theorem, \(\angle CAD=4^{\circ}\). Also, \(BC\) and \(BD\) are tangent to the circle (Earth), so \(\angle BCA=\angle BDA = 90^{\circ}\). In quadrilateral \(BCAD\), the sum of interior angles is \(360^{\circ}\). Let \(\angle BAD = x\). Then \(x + 90^{\circ}+90^{\circ}+4^{\circ}=360^{\circ}\), so \(x = 176^{\circ}\). But we can also use the formula for the angle formed by two tangents: \(\angle CBD=\frac{1}{2}(m\widehat{CED}-m\widehat{CD})\) (where \(CED\) is the major arc). But another way is to use the right - triangle relationship. We know that \(\cos\angle BAD=\frac{AD}{AB}\) (in right - triangle \(ABD\), \(AD\) is the radius of the Earth \(r = 4000\) mi, and \(AB=4000 + h\) where \(h\) is the height above sea - level). Also, \(\angle BAD=\frac{1}{2}(360^{\circ}-m\widehat{CD})\) (the central angle related to the angle formed by two tangents). So \(\angle BAD = 178^{\circ}\), and \(\angle ABD=\frac{1}{2}(m\widehat{CD})=2^{\circ}\) (using the property that the angle between a tangent and a radius is \(90^{\circ}\) and angle - sum in \(\triangle ABD\)).

Step2: Use the cosine function in right - triangle \(ABD\)

In right - triangle \(ABD\), \(\cos\angle ABD=\frac{AD}{AB}\). We know \(AD = 4000\) mi, \(\angle ABD = 2^{\circ}\), and \(AB=4000 + h\). So \(\cos(2^{\circ})=\frac{4000}{4000 + h}\). Then \(4000 + h=\frac{4000}{\cos(2^{\circ})}\).

Step3: Solve for \(h\)

We know that \(\cos(2^{\circ})\approx0.9994\). Then \(4000 + h=\frac{4000}{0.9994}\approx4002.40\). Subtract \(4000\) from both sides: \(h\approx4002.40−4000 = 2.40\)

Answer:

\(2.40\)