Elevation angles

From the endpoints of the base 240 m long and inclined at an angle of 18° 15 ', the top of the mountain can be seen at elevation angles of 43° and 51°. How high is the mountain?

Correct answer:

h =  1174.9592 m

Step-by-step explanation:

$$\begin{gathered} a=240\text{ m } \\ \alpha =18+15\mathrm{/}60={\displaystyle \frac{73}{4}}=18.25^{\circ } \\ {h}_1=a\cdot \sin (\alpha )=240\cdot \sin (18.25^{\circ })\doteq 75.1593\text{ m } \\ \cos \alpha =x_0:a \\ {x}_0=a\cdot \cos (\alpha )=240\cdot \cos (18.25^{\circ })\doteq 227.9278\text{ m } \\ \beta =43^{\circ } \\ \gamma =51^{\circ } \\ \tan \beta =(h-h_1):(x_0+x) \\ \tan \gamma =h:x \\ \tan \beta =(x\cdot \tan \gamma -h_1):(x_0+x) \\ (x_0+x)\cdot \tan \beta =(x\cdot \tan \gamma -h_1) \\ {x}_0\cdot \tan \beta +x\cdot \tan \beta =x\cdot \tan \gamma -h_1 \\ x={\displaystyle \frac{h_1+x_0\cdot \tan {\beta }^{\circ }}{\tan {\gamma }^{\circ }-\tan {\beta }^{\circ }}}={\displaystyle \frac{h_1+x_0\cdot \tan 43^{\circ }}{\tan 51^{\circ }-\tan 43^{\circ }}}={\displaystyle \frac{75.1593+227.9278\cdot \tan 43^{\circ }}{\tan 51^{\circ }-\tan 43^{\circ }}}={\displaystyle \frac{75.1593+227.9278\cdot 0.932515}{1.234897-0.932515}}=951.46321\text{ m } \\ h=x\cdot \tan {\gamma }^{\circ }=x\cdot \tan 51^{\circ }=951.4632\cdot \tan 51^{\circ }=951.4632\cdot 1.234897=1174.95922=1174.9592\text{ m } \\ \text{ Verifying Solution: } \\ {\beta }_1={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan ({\displaystyle \frac{h-h_1}{x_0+x}})={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan ({\displaystyle \frac{1174.9592-75.1593}{227.9278+951.4632}})=43^{\circ } \\ {\gamma }_1={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (h\mathrm{/}x)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (1174.9592\mathrm{/}951.4632)=51^{\circ } \end{gathered}$$

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