1. ## Triangle

need quick help! i dont understand these at all and i have a test tomorrow. damn summer school. anyways
A tourist walks n km at 4 km/h and then travels 2n km at 36 km/h by bus. Express the total traveling time t in hours as a function of n.
Answer is t(n) = 11n/36 i dont know how to get it though

A light 3m above the ground causes a boy 1.8m tall to cast a shadow s meteres long measured along the ground. express s as a function of d, the boy's distance in meters from the light. s(d) = 1.5d is the answer
Express the area A of a 30-60-90 right triangle as a function of the length of hypotenuse, h. answer is a(h)=h2root3/8

2. [quote=PrecalcKid114;59688]need quick help! i dont understand these at all and i have a test tomorrow. damn summer school. anyways
A tourist walks n km at 4 km/h and then travels 2n km at 36 km/h by bus. Express the total traveling time t in hours as a function of n.
Answer is t(n) = 11n/36 i dont know how to get it though[\quote]

recall that $\mbox { Speed } = \frac { \mbox { Distance }}{ \mbox { Time }}$

$\Rightarrow \mbox { Time } = \frac { \mbox { Distance }}{ \mbox { Speed }}$

Let $t_1$ be the time the journey on foot takes
Let $t_2$ be the time the journey by bus takes
Let $T$ be the total time

Thus we have $T = t_1 + t_2$

For $t_1$:

$t_1 = \frac {d}{s} = \frac {n}{4}$

For $t_2$:

$t_2 = \frac {d}{s} = \frac {2n}{36} = \frac {n}{18}$

Thus we have: $T = \frac {n}{4} + \frac {n}{18} = \frac {11n}{36}$

3. Originally Posted by PrecalcKid114
A light 3m above the ground causes a boy 1.8m tall to cast a shadow s meteres long measured along the ground. express s as a function of d, the boy's distance in meters from the light. s(d) = 1.5d is the answer
did you draw a diagram? we have similar triangles here.

Let $h_1$ be the height of the large triangle
Let $b_1$ be the base of the large triangle
Let $h_2$ be the height of the small triangle
Let $b_2$ be the base of the small triangle

since the triangles are similar, we have:

$\frac {h_1}{b_1} = \frac {h_2}{b_2}$

$\Rightarrow \frac {s + d}{3} = \frac {s}{1.8}$

$\Rightarrow s + d = 1.6666667s$

$\Rightarrow d = 0.6666667s$

$\Rightarrow \boxed { s = 1.5d }$

i rounded off a bit, 1.666666.... and 0.66666.... repeats forever

4. Originally Posted by PrecalcKid114
Express the area A of a 30-60-90 right triangle as a function of the length of hypotenuse, h. answer is a(h)=h2root3/8
see the diagram below.

Let $A$ be the area of the triangle
Let $a$ be the base of the triangle
Let $b$ be the height of the triangle
Let $h$ be the hypotenuse of the triangle

We will use the sine trig ratio

Since $\sin 30 = \frac {b}{h}$ we have $b = h \sin 30 = \frac {h}{2}$

Since $\sin 60 = \frac {a}{h}$ we have $a = h \sin 60 = \frac {h \sqrt {3}}{2}$

The area of a triangle is $\frac {1}{2} base \times height$

so we have: $A = \frac {1}{2} \cdot \frac {h}{2} \cdot \frac {h \sqrt {3}}{2}$

$\Rightarrow \boxed { A = \frac {h^2 \sqrt {3}}{8} }$

by the way, we use "^" to show powers, what you typed was h2root3, that is totally different from h^2 root3