Given a Partial side length and two angles find the height of a given figure.

Given a Partial side length and two angles find the height of a given figure

The height of Mount Rushmore = H   ?

The faces on Mount Rushmore are 60 feet tall: h = 60 feet

Head at a 48 degree angle of elevation: $$\alpha = 48 \ensurement{^{\circ}}$$

His chin at a 44.76 degree angle of elevation: $$\beta = 44.76 \ensurement{^{\circ}}$$

The distance to the mountain = d

Solution:

$$\tan{(\beta)}=\frac{H-h}{d} \qquad d=\frac{H-h}{\tan{(\beta)}} \quad (1)$$

$$\tan{(\alpha)}=\frac{H}{d} \qquad d = \frac{H}{\tan{(\alpha)}} \quad (2)$$

(1) = (2):

$$\frac{H-h}{\tan{(\beta)}} =\frac{H}{\tan{(\alpha)}}$$

$$\frac {\tan{(\beta)}}{\tan{(\alpha)}} =\frac {H-h} {H} = 1- \frac{h}{H}$$

$$\frac{h}{H} = 1 - \frac {\tan{(\beta)}} {\tan{(\alpha)}} =\frac {{\tan{(\alpha)}} -{\tan{(\beta)}} } {\tan{(\alpha)}} \quad | \quad \updownarrow$$

$$\frac {H} {h} =\frac {\tan{(\alpha)}} {{\tan{(\alpha)}} -{\tan{(\beta)}} }$$

$$\boxed{ H = h \left( \frac {\tan{(\alpha)}} {{\tan{(\alpha)}} -{\tan{(\beta)}} } \right) }$$

$$H = 60\left( \frac {\tan{(48\ensurement{^{\circ}})}} {{\tan{(48\ensurement{^{\circ}})}} -{\tan{(44.76\ensurement{^{\circ}})}} } \right)$$

H =  60 * 9.33639328817

H = 560.183597290 feet

The height of Mount Rushmore is 560.18 feet