tis question is so confusing.hu can solve?

**xiaoz** · June 19th 2006, 03:12 AM

wat is tis question toking about ?

**Quick** · June 19th 2006, 04:15 AM

Originally Posted by xiaoz

wat is tis question toking about ?

(a)The arc lamp to the shadow, and the man's head to the shadow form two similar triangles.
The ratio of height to base for the man's triangle is $\frac{1.8}{2.5}$
The ratio of height to base for the arc-lamp triangle is $\frac{x}{10+2.5}$
Because the triangles are similar, the two ratios are equal...
$\frac{1.8}{2.5}=\frac{x}{10+2.5}$

$\frac{1.8}{2.5}=\frac{x}{12.5}$

$\frac{1.8}{2.5}\cdot12.5=\frac{x}{1}$

$\frac{1.8\cdot12.5}{2.5}=x$

$\frac{22.5}{2.5}=x$

$9=x$

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: $(x_1,y_1)=(0,0)$

$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=d$

$\sqrt{\left(10-0\right)^2+\left(20-0\right)^2}=d$

$\sqrt{\left(10\right)^2+\left(20\right)^2}=d$

$\sqrt{100+400}=d$

$\sqrt{500}=d$

$22.3606797749979=d$

Just to make it easy to write, we'll say the distance between the man and the lamp is $\sqrt{500}$

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

$\frac{9}{\sqrt{500}+x}=\frac{1.8}{x}$

and solve from there.

**ticbol** · June 19th 2006, 05:09 AM

Originally Posted by xiaoz

wat is tis question toking about ?

i) On your drawing/figure as posted, draw a vertical line from the lamp to the ground.
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.

**xiaoz** · June 20th 2006, 12:32 AM

omg.. u all so clever.. but it is confusing ya...

**^_^Engineer_Adam^_^** · August 3rd 2006, 05:52 AM

W/ this equation
$<br /> \frac{9}{\sqrt{500}+x}=\frac{1.8}{x}<br />$
i have a question sir Quick
where is the x in the drawing?

**Quick** · August 3rd 2006, 07:01 AM

Originally Posted by ^_^Engineer_Adam^_^

W/ this equation
$<br /> \frac{9}{\sqrt{500}+x}=\frac{1.8}{x}<br />$
i have a question sir Quick
where is the x in the drawing?

x is the distance from the man's feet to the tip of the shadow (it's not the same x as the 1st problem)

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