Answer:
4713.23 ft
Explanation:
A sketch of the problem is given below:
• The survey team was initially at point A above, the angle of elevation is 26 degrees.
,• They moved 1500 feet closer to point B and the angle of elevation is 30 degrees.
We want to find the height, h of the mountain.
Recall from trigonometry:
[tex]\tan \theta=\frac{\text{Opposite}}{\text{Adjacent}}[/tex]In right triangle EFB:
[tex]\begin{gathered} \tan B=\frac{EF}{FB} \\ \tan 30\degree=\frac{h}{x} \\ \implies h=x\tan 30\degree \end{gathered}[/tex]Likewise, in right triangle EFA:
[tex]\begin{gathered} \tan A=\frac{EF}{FA} \\ \tan 26\degree=\frac{h}{x+1500} \\ \implies h=(x+1500)\tan 26\degree \end{gathered}[/tex]Equate the heights from the two obtained above:
[tex]x\tan 30\degree=(x+1500)\tan 26\degree[/tex]Then, solve the equation for x.
[tex]\begin{gathered} x\tan 30\degree=(x+1500)\tan 26\degree \\ \text{Open the bracket on the right-hand side} \\ x\tan 30\degree=x\tan 26\degree+1500\tan 26\degree \\ \text{Subtract }x\tan 26\degree\text{ from both sides.} \\ x\tan 30\degree-x\tan 26\degree=x\tan 26\degree-x\tan 26\degree+1500\tan 26\degree \\ x\tan 30\degree-x\tan 26\degree=1500\tan 26\degree \\ \text{Factor out x} \\ x(\tan 30\degree-\tan 26\degree)=1500\tan 26\degree \\ \text{Divide both sides by }(\tan 30\degree-\tan 26\degree) \\ \frac{x(\tan30\degree-\tan26\degree)}{(\tan30\degree-\tan26\degree)}=\frac{1500\tan26\degree}{(\tan30\degree-\tan26\degree)} \\ \\ x=8163.5552ft \end{gathered}[/tex]Finally, substitute x=8163.56 ft to solve for h, the height.
[tex]\begin{gathered} h=x\tan 30\degree \\ =8163.5552\times\tan 30\degree \\ =4713.23ft \end{gathered}[/tex]The height of the mountain is 4713.23 ft correct to 2 decimal places.
