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Gnom [1K]
3 years ago
12

In sunlight, a vertical yardstick casts a 1 ft shadow at the same time that a nearby tree casts a 15 ft shadow. How tall is the

tree?
Mathematics
2 answers:
nadya68 [22]3 years ago
7 0

To answer this specific problem, the tree is 45 feet tall. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

joja [24]3 years ago
3 0

Answer:

Tree is 45 ft tall.

Step-by-step explanation:

In the figure attached AB is the tree and CD is the yardstick.

Shadow of tree AB is BO = 15 ft and shadow of yardstick is CO.

since length of a yardstick is = 1 yard or 3 ft.

Now we know that ΔABO and ΔDCO are similar.(since ∠A = ∠D and ∠B = ∠C = 90° so AA property proving triangles similar)

Therefore \frac{AB}{CD}=\frac{OB}{OC}

By putting the values of AB and CD

\frac{x}{3}=\frac{15}{1}

x = 3×15 = 45 ft.

Therefore the height of the tree is 45 ft.

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On a morning of a day when the sun will pass directly​ overhead, the shadow of a 60​-ft building on level ground is 25 ft long.
AysviL [449]

Answer:

The shadow is decreasing at the rate of 3.55 inch/min

Step-by-step explanation:

The height of the building = 60ft

The shadow of the building on the level ground is 25ft long

Ѳ is increasing at the rate of 0.24°/min

Using SOHCAHTOA,

Tan Ѳ = opposite/ adjacent

= height of the building / length of the shadow

Tan Ѳ = h/x

X= h/tan Ѳ

Recall that tan Ѳ = sin Ѳ/cos Ѳ

X= h/x (sin Ѳ/cos Ѳ)

Differentiate with respect to t

dx/dt = (-h/sin²Ѳ)dѲ/dt

When x= 25ft

tanѲ = h/x

= 60/25

Ѳ= tan^-1(60/25)

= 67.38°

dѲ/dt= 0.24°/min

Convert the height in ft to inches

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Therefore, 60ft = 60*12

= 720 inches

Convert degree/min to radian/min

1°= 0.0175radian

Therefore, 0.24° = 0.24 * 0.0175

= 0.0042 radian/min

Recall that

dx/dt = (-h/sin²Ѳ)dѲ/dt

= (-720/sin²(67.38))*0.0042

= (-720/0.8521)*0.0042

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Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min

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3 years ago
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