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From a point on the ground the angle of elevation of the top of a tower is x°. Moving 150 meters away from that point the angle of elevation was found to be y° .If tan x=3/4 and tan y=5/7 find the height of the tower.​

From a point on the ground the angle of elevation of the top of a tower is x°. Moving-example-1

2 Answers

4 votes

Answer:

2250m

Explanation:


tangent=(opposite)/(adjacent)

We have:


\tan x^o=(3)/(4)\\\\\tan y^o=(5)/(7)

By definition of tangent, we have:


\tan x^o=(AB)/(AC)\\\\\tan y^o=(AB)/(AC+150)

Therefore we have the system of equations:


\left\{\begin{array}{ccc}(AB)/(AC)=(3)/(4)&(1)\\\\(AB)/(AC+160)=(5)/(7)&(2)\end{array}\right

From (1)


(AB)/(AC)=(3)/(4) cross multiply


3AC=4AB divide both sides by 3


AC=(4AB)/(3)

Substitute it to (2):


(AB)/((4AB)/(3)+150)=(5)/(7)\\\\(AB)/((4AB)/(3)+(3\cdot150)/(3))=(5)/(7)\\\\(AB)/((4AB)/(3)+(450)/(3))=(5)/(7)\\\\(AB)/((4AB+450)/(3))=(5)/(7)\\\\AB\cdot(3)/(4AB+450)=(5)/(7)


(3AB)/(4AB+450)=(5)/(7) cross multiply


(3AB)(7)=(5)(4AB+450)\\\\21AB=(5)(4AB)+(5)(450)


21AB=20AB+2250 subtract 20AB from both sides


AB=2250

Such a tower height is rather impossible, but this is the solution.

User Merk
by
5.7k points
3 votes

Answer:

2250m

Explanation:

Step 1

Since

Hence,

tan x = 3/4

tan y=5/7

tan x = BA/ CA where BA = height and CA = distance

3/4 = h/ d

4h = 3d

h = 3d/4.......... Equation 1

tan y = BA / DA + 150m

5/7 = h/d + 150

7 × h = 5(d + 150)

7h = 5d + 750............ Equation 2

Since h = 3d/4

7(3d/4) = 5d + 750

21d/4 = 5d + 750

Multiply both sides by 4

21d = 4(5d + 750)

21d = 20d + 3000

21d -20d = 3000

d = 3000

Distance (d) = 3000m

Substitute 3000m for d in equation 1

h = 3d/4

h = 3 × 3000/4

h = 2250m

User Priyath Gregory
by
4.8k points