Optimizing Angle for Shooting at a Distance with a Sniper Rifle

[taffman123]

what was my mistake

 

[Hootenanny]

Multiplying fractions:
a \times \frac{b}{c} = \frac{a}{1} \times \frac{b}{c} = \frac{a \cdot b}{c}
So your first term should be:
\frac{1500 \times 854 \cdot \sin x}{854 \cos x} = \frac{1500 \cdot \sin x}{\cos x}

 

[taffman123]

question: where does the t come from - did we not sub in for t?

 

[Hootenanny]

Again, an oversight on my part,must be getting tired.

 

[taffman123]

ok so we have

1500sinx/cosx - 4.9(1500/854cosx)^2 - 40

1500tanx ( correct?) -4.9 (3.085cosx^2) -40 = 0

1500 tanx - 15.1169cos^2x-40 =0

is that correct?

 

[taffman123]

but i still don't see how that helps me find x?

 

[Hootenanny]

Your are correct. You need to get the two trig functions in the same form (sin, cosine, sec, etc.), but at the moment I can't recall any useful trig identities.

 

[taffman123]

i only have learned 2

sinx/cosx = tanx

and sin2x + cos2x = 1

 

[Hootenanny]

It may be worth venturing into the Math forums and asking about the idents.

 

[taffman123]

have u been able to figure out the answer

or would u be able to explain what hadeijv did ? because he is finished

 

[Hootenanny]

hadeijv squared the whole equation, giving as topsquark says a bi-quadratic. This allowed himto use the ident \sin^2 x + \cos^2 x = 1 to remove the sin and just leave cosine function.

 

[taffman123]

can i do that in my case as well?

 

[Hootenanny]

Yes, you are both working on the same problem and using the same equations. Looking at the other thread may give you some hints. Topsquark explained it well.

 

[MrRottenTreats]

okay, I am trying to work this out I am at 1500sinx / cosx -15/cos2x - 40 = 0 ... and i do not know where to go form here. and is this correct?