A survey team is trying....

Question

1. ) A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that the angle of elevation to the top of the mountain is 24degrees . From a point 1500 feet closer to the mountain along the plain, they find that the angle of elevation is 29 degrees. How high (in feet) is the mountain? 
2.  A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that the angle of elevation to the top of the mountain is 29 degrees . From a point 1000 feet closer to the mountain along the plain, they find that the angle of elevation is 32 degrees How high (in feet) is the mountain? 

Answer 1

DRAW A PICTURE!!!

mountain on the left...

top of the mountain is point A.

Base of the mountain directly below touches the ground at point B.

Point C is the point where the surveyors were ORIGINALLY STANDING.

Triangle ABC is a right triangle, where AB is the height and BC is the base.

M is the point in between the base of the mountain, point B, to where the surveyors

WERE ORIGINALLY standing, point C.

y = BM and x = MC

So x is the distance they move closer to the base of the mountain.

So therefore, x+y = BC = BM+MC

angle theta = angle T = angle CMA

angle lambda = angle L = angle MBA

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tan T = h/(x+y)

tan L = h/y

y * tan L = h <--- Please label this equation ALPHA

tan T = (y * tanL)/(x+ y)

tan T ( x+y) = y*tanL

tanT* x + tanT * y = y * Tan L

tan T*x = y * Tan L - y * tan T

tan T * x = y ( tan L - tan T)

tan T * x/ (tan L - tan T) = y

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T = 24, x = 1500, L = 29

y = tan 24*1500/ ( tan 29 - tan 24) = 6122.48612256894345-

Per equation ALPHA, the height is just under 3393.75

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T=29, x=1000, L = 32

y = tan 29*1000 / (tan 32 - tan 29) = 7855.82+

Per equation ALPHA the height is 4908.86143+

Answer 2

Since angle BCD is 29°, then angle BCA is 151 (supplementary angles). Since triangle ABC has 180°, we can calculate angle ABC as 5°.

Now we can use the Law of Sines to find side x. sin(5°)/1500 = sin(24°)/x so x = 1500*sin(24°)/sin(5°) = 7000.169.

Now using triangle BCD (a right triangle) we can say sin(29°) = h/7000.169 so h = 7000.169*sin(29°) = 3394 ft.

The second problem is the same, just different numbers.

Answer 3

For question 1, the first time they observe the angle, we can say that they are a distance x away from the mountain, with an angle of elevation of 24 degrees so we can make a triangle with angle of elevation of 24, length x on the horizontal and height h on the vertical, with a right angle between the horizontal and vertical. Therefore, we have tan(24) = h/x. The second time they observe, we have an angle of elevation of 29 degrees, a length of x-1500 on the horizontal, and height h on the vertical, and a right angle between the horizontal and vertical. Therefore, we have tan(29) = h/(x-1500). Using these two equations, we can solve for h and get h = xtan(24) from the first equation, and h = (x-1500)tan(29). Setting these equal to each other, we get xtan(24) = (x-1500)tan(29). From this we can put all the terms with x on the same side and get 1500tan(29) = x(tan(29)-tan(24)). Solving for x, we get x = 1500tan(29)/(tan(29)-tan(24)). Plugging this into h = xtan(24), we get h = 1500tan(29)tan(24)/(tan(29)-tan(24)).

For question 2, applying the same process we get h = 1000tan(32)tan(29)/(tan(32)-tan(29)).

Hope this helps! Let me know if there is anything I can clarify!