Question

From a point on the ground the angles of elevation of the bottom and top of a water tank kept on the top of the 30 metre high building are 30 degree and 45 degree respectively find the height of the water tank.

can anyone please give me the right answer .???
It's really urgent mate..!!!
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Answer 1

If a tank is kept at the top of a building of height 30 m and the angle of elevation of bottom & top of tank is 30° & 45° then the height of the water tank is 30(√3-1) m or 21.96 m.

Step-by-step explanation:

The height of building = 30 m

The angle of elevation from the ground to the bottom of the tank = 30°

The angle of elevation from the ground to the top of the tank = 45°

Let the height of tank CD be “h” m and the distance AB be “x” m (as shown in the figure)

Considering ∆ ADB, we get

tan θ = perpendicular/base = AD/AB

⇒ tan 30 = 30/x ….. [here θ = 30°]

⇒ 1/√3 = 30/x

⇒ x = AB = 30√3 m …… (i)

Now, considering ∆ ABC,  

tan θ = AC/AB

⇒ tan 45 = (h+30)/x ….. [here θ= 45°]

⇒ 1 = (h+30)/30√3 …… [substituting the value of x from (i)]

⇒ h = 30√3 – 30

⇒ h = 30[√3 - 1] m or 21.96 m ….. [√3 = 1.732]  

Thus, the height of the tank is 30[√3 - 1] m or 21.96 m .

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Also View:

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Answer 2

Answer:

$30(\sqrt3-1) m$

Step-by-step explanation:

We are given that

Height of building= 30 m

Let Height of water tank=AB=h

BC= 30 m

In triangle BCD,

$\frac{BC}{CD}=tan30^{\circ}$

$tan\theta=\frac{perpendicular\;side}{base}$

$\frac{30}{CD}=\frac{1}{\sqrt3}$

$tan 30^{\circ}=\frac{1}{\sqrt3}$

$CD=30\sqrt3 m$

In triangle  ACD,

$\frac{AC}{CD}=tan45^{\circ}$

$\frac{AC}{CD}=1$

$tan45^{\circ}=1$

$AC=CD=30\sqrt3$

$AB+BC=30\sqrt3$

$h+30=30\sqrt3$

$h=30\sqrt3-30=30(\sqrt3-1)$

Hence, the height of water tank =$30(\sqrt3-1) m$