Elevation angles

Two endpoints distant 240 m are inclined at an angle of 18°15'. The top of the mountain can be seen at elevation angles of 43° and 51° from its. How high is the mountain?

Final Answer:

h =  1174.9592 m

Step-by-step explanation:

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

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