a survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain they observed at the angle of elevation to the top of the mountain is 26 degrees. From a point 1500 feet closer to the mountain along the plain, they find that the angle of elevation is 30 degrees. How high in feet is the mountain?

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03-11-2022
Mathematics

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a survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain they observed at the angle of elevation to the top of the mountain is 26 degrees. From a point 1500 feet closer to the mountain along the plain, they find that the angle of elevation is 30 degrees. How high in feet is the mountain?

Sagot :

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ElowynT501355 ElowynT501355 · Answer 1 · 2022-11-03T00:15:12-04:00

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.

Sagot :

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