Slope and Projectile Problem: Finding Distance Covered on a Landing Slope

[Lolagoeslala]

Okay so what i was thinking was i could find the side that is opposite to the angle
so like

tan 37 = o / 61.22 m
o = 46.13 m

and use a^2 = b^2 + c^2
a^2 = (61.22m)^2 + (46.13m)^2
a= 76.6 or 77

 

[gneill]

Okay so what i was thinking was i could find the side that is opposite to the angle
so like

tan 37 = o / 61.22 m
o = 46.13 m

and use a^2 = b^2 + c^2
a^2 = (61.22m)^2 + (46.13m)^2
a= 76.6 or 77

That'll do it. But it's kind of taking the long way around

You can write $cos(\theta) = x/L$ so that $L = x/cos(\theta)$, and cos(θ) can be obtained from the similar 3-4-5 triangle as cos(θ) = 4/5.

Or, avoiding trig functions altogether, you can use the "raw geometry" of similar triangles:

$\frac{L}{x} = \frac{5}{4}$ so that $L = x \frac{5}{4}$

where x is 61.18 m, of course.

 

[Lolagoeslala]

THANK YOU SO MUCH FOR YOUR HELP!
u r like a god to me :D

 

[gneill]

THANK YOU SO MUCH FOR YOUR HELP!
u r like a god to me :D

Careful, my ego may explode

Glad I could help. Cheers!