Answer to Question #172070 in Geometry for Derby

Question

Sighting from a helicopter: A helicopter hovers at an altitude that is 1000 feet above a mountain peak of altitude 5210 feet. A second, taller peak is viewed from both the mountaintop and the helicopter. From the helicopter, the angle of depression is 43 degrees,and from the mountaintop, the angle of elevation is 18 degree . How far apart are the mountain peaks? What is the altitude of the taller peak?

Answer 1

$\text{Let A,B be mountain peaks}$

$\text{C be helicopter's location}$

$\text{let B be such a point that:}$

$AD \, \text{is the distance between the mountain peaks}$

$BD \,\text{is the difference in height between mountain peaks}$

$\angle{BDA}=90\degree$

$\angle{CAD}=90\degree$

$\text{from the conditions of the problem}$

$\angle{BAD}=18\degree$

$\angle{ACB}=43\degree$

$CA = 1000$

$\text{find angles}$

$\alpha=\angle{CAB}= \angle{CAD}-\angle{BAD}=90\degree-18\degree=72\degree$

$\gamma=\angle{ACB}= 43^{\circ}$

$\beta=\angle{CBA}=180\degree-\angle{CAB}-\angle{BCA}=$

$180\degree-(72\degree+43\degree)=65\degree$

$\text {by the sine theorem }\triangle{ABC}$

$\frac{AB}{\sin{\gamma}}=\frac{AC}{\sin{\beta}}$

$AB=\frac{AC}{\sin{\beta}}\sin{\gamma}$

$\text{from } \triangle{ABD}$

$BD=AB*\sin{\angle{BAD}}=\frac{AC}{\sin{\beta}}\sin{\gamma}*\sin\angle{BAD}$

$BD = \frac{1000}{\sin65\degree}\sin43\degree\sin18\degree\approx$ 232.5

$AD=AB*\cos{\angle{BAD}}=\frac{AC}{\sin{\beta}}\sin{\gamma}*\cos\angle{BAD}$

$AD = \frac{1000}{\sin65\degree}\sin43\degree\cos18\degree\approx715.7$

$\text {then the height of the highest mountain peak}$

$h = BD+5210= 232.5+5210= 5442.5\ ft$

Answer: 715.7 ft distance between mountain peaks

5442.5 ft the height of the highest mountain peak