frozenmelody18
contestada

A motorcycle travels 20 miles west and then turns and goes 52° north of west another 38 miles. How far is the motorcycle from its starting point?? Round your answer to the nearest tenth
__ miles

Respuesta :

tanweeleong195oxyqwp

Answer:

52.7 miles

Step-by-step explanation:

Please refer to attached photo. (Apologies for the terrible drawing).

We can deduce from A to B the distance is 20 miles, therefore no calculations is required.

However, from B to C, we will have to find the horizontal and vertical distances.

Horizontal Distance B > C =  = hBC = [tex]BCcos(52)[/tex] = [tex]38cos(52) miles[/tex]

Vertical Distance B > C = vBC = [tex]BCsin(52)[/tex] = [tex]38sin(52)miles[/tex]

Here you can see the whole diagram is a right angle triangle. Which means, we can use Pythagoras' Theorem to find AC.

By Pythagoras Theorem,

[tex]c^{2} =a^{2}+b^{2}[/tex]

[tex]AC^{2} = (AB + hBC)^{2} + vBC^{2} \\AC^{2} = (20+38cos(52))^{2} +(38sin(52))^{2} \\AC = \sqrt{(20+38cos(52))^{2} +(38sin(52))^{2} } \\AC = 52.7 miles[/tex](nearest tenth)

Ver imagen tanweeleong195oxyqwp
BasketballEinstein

Answer:

[tex]\displaystyle 52,7\:miles[/tex]

Step-by-step explanation:

I have not experienced much with this topic while I was in school, but I do know a trick to this, so pay attention:

First, let us pretend that this triangle is a right triangle and use the Pythagorean Theorem to define the displacement [how far the motorcycle is from its starting point]:

[tex]\displaystyle a^2 + b^2 = c^2 \\ 20^2 + 38^2 = c^2 \\ \sqrt{1844} = \sqrt{c^2} \\ 2\sqrt{461}\:[or\:42,941821107...] = c[/tex]

Knowing that this is an obtuse triangle, we will have to integrate and use a bit of physics. So, think back to calculating potential energy, [tex]\displaystyle mgh = PE.[/tex] The g represents acceleration due to gravity, which is at approximately [tex]\displaystyle 9\frac{4}{5}\:\frac{m}{s}^2,[/tex] which obviously rounds to [tex]\displaystyle 10\:\frac{m}{s}^2[/tex], but since you need it rounded to the nearest tenth, we will add [tex]\displaystyle 9\frac{4}{5}[/tex] to [tex]\displaystyle 2\sqrt{461},[/tex] giving you this:

[tex]\displaystyle \boxed{\frac{49 + 10\sqrt{461}}{5}} = 2\sqrt{461} + 9\frac{4}{5} \\ \boxed{52,741821107...} = 42,941821107... + 9,8[/tex]


So, when rounded to the nearest tenth, the motorcycle is approximately [tex]\displaystyle 52,7\:miles[/tex] from its starting position.


I am joyous to assist you at any time.