When you are given a right triangle and you are able to find all of the side
lengths, how do you find the last two angles?
If you have all the sides, then all you need to do is pick one of the acute
angles and call it theta.
From the position of that angle, identify the opposite side and the adjacent
side.
Then you have
tan(theta) = opp / adj.
You know opp and adj, so opp / adj is just a number.
Take the arctan of both sides (2d function - tan). You get
arcTan(tan(theta)) = arcTan(opp / adj) so theta = arcTan(opp / adj).
Be sure you are in degree mode if you want theta in degrees, radian mode if
you want the angle in radians.
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Question: How to find the six function given the following information.
Homework Problem # 114 pg 444
What I know: Angle theta is given Opposite side is given as: 1/q Hypotenuse
is given as q These two will give us the sin theta: 1/q over q solving a2 + b2 =
c2 is where I am running into a problem. Making adjacent side x I have (1/q)2 +
x2 = q2 I subtract (1/q)2 from both sides
You're doing great up to here, but you aren't doing the right thing at this
point.
You have
(1/q)^2 + x^2 = q^2, which can be written 1 / q^2 + x^2 = q^2.
Whenever you have an equation with fractions, your the first thing you want
to do is multiply every term by the common denominator. In this case, that is
q^2.
So you have
q^2 ( 1 / q^2) + q^2 * x^2 = q^2 * q^2, or 1 + q^2 x^2 = q^4.
We want to solve for x. So subtract 1 from both sides to get
q^2 x^2 = q^4 - 1. Then divide through by q^2 to get x^2 = (q^4 - 1) / q^2.
Finally take the positive square root of both sides to get x = sqrt( q^4 - 1) /
q.
This will be a real number as long as q >= 1.
If q < 1 then 1 / q > 1 so 1/q > q. If q is the hypotenuse, this can't be.
The opposite side can't be longer than the hypotenuse.
So for any possible triangle, you do have a solution.
I take the square root of both sides and after finding the common denominator
I end up with a negative number. I know a side of a triangle can not have a
negative measurement.
What I need to know: If I am messing up the a2 + b2 = c2 equation when I am
solving it. Or if I need to be looking at this problem in a different way to
solve it
For some reason, I am not able to find the csc sec or cot of a angle with my
calculator (it says an error has occured) What am I doing wrong? The regular
functions work just fine.
I can't tell you for sure what the error might be. Most calculators don't
even have csc or sec or cot buttons, and you have to take reciprocals of sines,
cosines and tangents to find these functions. So you can use this method until
you resolve your original question.
Exactly what sequence of keys do you use to find the tangent of an angle?
Exactly what sequence of keys are you trying to use to find the csc of an angle?
If the set of angles whose measures are greater than 0 and less than 90 is
domain for each function, What appears to be the ranges for sine, the cosine,
and the tangent functions given this domain? #33 in section 5.2
Visualize a series of right triangles, with theta increasing from 0 deg to
90 deg while the hypotenuse stays the same. Does the opposite side increase or
decrease?
So does sin(theta) increase or decrease when theta increases from 0 deg to
90 deg?
What are sin(0 deg) and sin(90 deg)?
What do you therefore think is the range of the sine function for 0 <= theta
<= 90 deg?
Visualize the same series right triangles, but think about the adjacent
side. Does it increase or decrease?
So does cos(theta) increase or decrease when theta increases from 0 deg to
90 deg?
What are cos(0 deg) and cos(90 deg)?
What do you therefore think is the range of the cosine function for 0 <=
theta <= 90 deg?
Finally thing about the tangent, which is opp / adj. Imagine a series of
right triangles, all having the same adjacent side. In order for the angle to
increase from 0 to 90 deg, what has to happen to the opposite side?
Is the tangent therefore increasing or decreasing?
Is there any limit to how big the opposite side gets as theta approaches 90
deg?
What is tan(0 deg)? What happens to the value of the tangent as theta
approaches 90 deg?
What do you therefore think is the range of the tangent function for 0 <=
theta <= 90 deg?
Will you give us credit if we are around the right estimate when we eyeball
the triangles to find the info.? I know what sin, cos, etc. are but it is hard
to eyeball some of the triangles.
Any reasonable estimate is acceptable. I'm not going to split hairs, as long
as you aren't making a clear error.