Question

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.​

Answer 1

Answer:

2250m

Step-by-step 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

Answer 2

Answer:

2250m

Step-by-step explanation:

$tangent=\dfrac{opposite}{adjacent}$

We have:

$\tan x^o=\dfrac{3}{4}\\\\\tan y^o=\dfrac{5}{7}$

By definition of tangent, we have:

$\tan x^o=\dfrac{AB}{AC}\\\\\tan y^o=\dfrac{AB}{AC+150}$

Therefore we have the system of equations:

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

From (1)

$\dfrac{AB}{AC}=\dfrac{3}{4}$             cross multiply

$3AC=4AB$         divide both sides by 3

$AC=\dfrac{4AB}{3}$

Substitute it to (2):

$\dfrac{AB}{\frac{4AB}{3}+150}=\dfrac{5}{7}\\\\\dfrac{AB}{\frac{4AB}{3}+\frac{3\cdot150}{3}}=\dfrac{5}{7}\\\\\dfrac{AB}{\frac{4AB}{3}+\frac{450}{3}}=\dfrac{5}{7}\\\\\dfrac{AB}{\frac{4AB+450}{3}}=\dfrac{5}{7}\\\\AB\cdot\dfrac{3}{4AB+450}=\dfrac{5}{7}$

$\dfrac{3AB}{4AB+450}=\dfrac{5}{7}$            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.