# Math Help - Word Problems

1. ## Word Problems

a.) A water tank has the shape of a cone (like an ice cream cone without ice cream). The tank is
10m high and has a radius of 3m at the top. If the water is 5m deep (in the middle) what is
the surface area of the water?

b.) A kite is 100m above the ground. If there are 200m of string out what is the angle between the
string and the horizontal. (Assume that the string is perfectly straight.)

2. Originally Posted by saar4ever
a.)
b.) A kite is 100m above the ground. If there are 200m of string out what is the angle between the
string and the horizontal. (Assume that the string is perfectly straight.)

vertical = 100 m , hypothenus = 200 m

sin a = 100/200

a=arcsin 1/2

a=30 degrees .

3. thanks just got a.) left its pretty hard been trying it for over 2 hours dunno how to find radius at halfway the cone is it simply 1.5 radius? or more complicated?
if it is 1.5 radius then pie r squared is 3.1415 times 1.5 squared which is = to 7.069 is that correct surface area for a.)?

4. ## Part b

part b is easy. using your soh cah toa rule, u can use the soh part which is sinθ=o/h where o=opposite which is 100 and h=hypotenuse which equals 200 since there is 200m of string. therefore;
sinθ=o/h
sinθ=100/200
sinθ=0.5
θ=sin^(-1)x0.5
i'll let you figure out the answer:P

5. Originally Posted by saar4ever
a.) A water tank has the shape of a cone (like an ice cream cone without ice cream). The tank is
10m high and has a radius of 3m at the top. If the water is 5m deep (in the middle) what is
the surface area of the water?

first of all draw a cone , then you will see clearly . Now note that the radius of the water, r is different from the radius on top of the cone .

By similar triangles , 3/10=r/5 .... r=3/2

so the surface area of the water = $\pi(\frac{3}{2})^2+\pi(\frac{3}{2})l$ ,
you can find the l using phythagoras .

$\pi(\frac{3}{2})^2$
$+\pi(\frac{3}{2})l$
does the question ask for the side area of water?
if yes then ur version is correct
31.668 right?

7. Originally Posted by saar4ever
$\pi(\frac{3}{2})^2$
$+\pi(\frac{3}{2})l$
$\pi r^2$ is the top part of the water and $\pi rl$ is the curved surface area of the water . I don see why this part don exist .
$\pi rl$ is the curved surface area of the water . I don see why this part don exist .