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#1




Help me solve a math problem
This is not homework. Pinky swear.
I love to try word problems that come my way. This one simplifies to 300cos(angle) = 5sin(angle) + 3 It works out to around 88.5 degrees but I was only able to solve it by trial and error. I would appreciate a trig lesson here. Thanks. 
#2




If you square both sides, then apply sin^2 + cos^2 = 1, it becomes a quadratic equation, which you can solve any number of ways but the quadratic formula is probably easiest.
90000cos^2 = 25sin^2 + 30sin + 9 90000(1sin^2) = 25sin^2 + 30sin + 9 Then substitute x=sin^2 and solve for x Last edited by sbunny8; 02162018 at 12:24 PM. 
#3




I'm curious just what word problem you were doing that ended up reducing to that.
And it should be noted that, although this one is solvable, it's very easy when putting together equations involving sines and cosines to end up with what's called a transcendental equation, for which the only recourse is something like your trialanderror solution (though of course, there are usually more efficient methods than pure trialanderror available). 
#4




Another method is to apply trig identities. Your equation is
300 cos x  5 sin x = 3. Now draw a right triangle with legs 300 and 5. The hypotenuse of this triangle will be q = √(300^{2} + 5^{2}). The lefthand side of your equation can be rewritten as q ( (300/q) cos theta  (5/q) sin alpha) = 3. But if you go back to your drawing of the triangle, and you define alpha to be the angle between the length300 side and the hypotenuse, then you can see that 300/q = cos alpha and 5/q = sin alpha. Thus, this is equivalent to q [(cos alpha)(cos theta)  (sin alpha)(sin theta)] = 3. The quantity in square brackets is a trigonometric addition identity, meaning that cos (alpha + theta) = 3/q. Thus, theta = cos^{1} (3/q)  alpha, with alpha and q defined as above. This sort of trick can be used in general to solve for an angle that appears in both a sine and a cosine in an equation. Last edited by MikeS; 02162018 at 03:27 PM. 


#5




Quote:
Here is the word problem as I remember. A river is 0.5 km wide. A boat must cross the river to a point 15 km downstream. The current is 12 km/h and the speed of the boat is 20 km/h. At what angle must the boat head? My solution involved breaking it up into components and having the t for the vertical and horizontal motions be equal. There must a vector solution as well. I was able to solve the quadratic that resulted from the hint of squaring as above. I got 88.9 degrees which I found is not correct through a simple check. Getting closer but still stumped. 
#7




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#8




A second solution is Θ=89.618º

#9




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