Question

The angular elevation of a tower CD at a place A due south of it is 60 and at a place B due west of A, the elevation is 30. If AB =3km, then find the height of the tower.

Answer 1

I hope this will help you

Answer 2

If AB =3km, then the height of the tower is

$\frac{3\sqrt{6}}{4} km$.

Step-by-step explanation:

Let the height of the tower CD be "h".

Referring to the figure attached below, we will further solve the given question:

Step 1:

Consider right-angled Δ ACD, applying right angle trigonometry ratios, we get

tan θ = perpendicular/base = CD/AC

⇒ tan 60° =  h/AC

⇒ √3 = h/AC

⇒ AC = h/√3 ....... (i)

Step 2:

Consider right-angled Δ BCD, applying right angle trigonometry ratios, we get

tan θ = perpendicular/base = CD/BC

⇒ tan 30° =  h/BC

⇒ 1/√3 = h/BC

⇒ BC = h√3 ....... (ii)

Step 3:

Now, considering right-angled ΔBAC, applying Pythagoras Theorem,we get

BC² = AB² + AC²

⇒ (h√3)² = (3)² + (h/√3)² ....... [substituting values of BC & AC from (i) & (ii)]

⇒ 3h² = 9 + h²/3

⇒ 3h² - h²/3 = 9

⇒ 9h² - h² = 9*3

⇒ 8h² = 27

⇒ h² = 27/8

⇒ h = $\frac{3\sqrt{3} }{2\sqrt{2} }$

⇒ h = $\frac{3\sqrt{3} }{2} * \frac{\sqrt{2}}{\sqrt{2}}$

⇒ h = $\frac{3\sqrt{6} }{4}$ km

Thus, the height of the tower is $\frac{3\sqrt{6} }{4}$ km.

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