wat is tis question toking about ?

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- June 19th 2006, 04:12 AMxiaoztis question is so confusing.hu can solve?
wat is tis question toking about ?

- June 19th 2006, 05:15 AMQuickQuote:

Originally Posted by**xiaoz**

The ratio of height to base for the man's triangle is

The ratio of height to base for the arc-lamp triangle is

Because the triangles are similar, the two ratios are equal...

Which means the height of the arc-lamp triangle is 9m.

(b) This is an interesting question. Imagine the man on a co-ordinate plane with the arc-lamp in the center. He is on point (10,0) and walks to point (10,20).

We need to first find the distance between him and the lamp. The distance formula is...

note:

Just to make it easy to write, we'll say the distance between the man and the lamp is

Now we know the heights of both triangles, we know they're similar, so we remake our ratios...

and solve from there. - June 19th 2006, 06:09 AMticbolQuote:

Originally Posted by**xiaoz**

Let angle BCA = theta.

In the smaller right triangle,

tan(theta) = 1.8/2.5

In the larger right triangle, let x = height of lamp,

tan(theta) = x/(10+2.5)

tan(theta) = tan(theta),

1.8/2.5 = x/(12.5)

x = (1.8/2.5)(12.5) = 9m ------------answer.

-------------------

ii) If the man walks south, you cannot see it on your posted figure.

Go above and over the lamp. Look down. Walk the man now 20m due south. You have a new right triangle. The position now of the man from the base of the lamp post is, by Pythagorean Theorem,

d^2 = (10)^2 +(20)^2

d^2 = 500

d = 10sqrt(5) m.

Now go down to the ground again. Look at the man and the lamp post now. You have two new right triangles again.

Let y = length of the man's shadow now.

New smaller right triangle:

---hypotenuse = ray of light from the lamp = unknown

---vertical leg = man's height = 1.8m

---horizontal leg = new shadow = y

---angle between hypotenuse and horizontal leg = say, alpha.

New larger right triangle:

---hypotenuse = ray of light from the lamp = unknown

---vertical leg = lamp's height = 9m

---horizontal leg = (10sqrt(5) +y)m

---angle between hypotenuse and horizontal leg = alpha also.

So, in smaller triangle,

tan(alpha) = 1.8 /y

In larger triangle,

tan(alpha) = 9 /(10sqrt(5) +y)

tan(alpha) = tan(alpha)

1.8 /y = 9 /(10sqrt(5) +y)

Cross multiply,

1.8(10sqrt(5) +y) = 9y

Divide both sides by 9,

2sqrt(5) +0.2y = y

2sqrt(5) = 0.8y

y = 2sqrt(5) /(0.8)

y = 2.5sqrt(5) = 5.59 m -------man's shadow now, answer. - June 20th 2006, 01:32 AMxiaoz
omg.. u all so clever.. but it is confusing ya... :eek:

- August 3rd 2006, 06:52 AM^_^Engineer_Adam^_^
W/ this equation

i have a question sir Quick

where is the x in the drawing? - August 3rd 2006, 08:01 AMQuickQuote:

Originally Posted by**^_^Engineer_Adam^_^**