SOLUTION: Find the slope of the line bisecting the angle from line1, with slope 2, to line2, with no slope. Hope you'll help me. Thanks! Will wait asap.

Algebra ->  Trigonometry-basics -> SOLUTION: Find the slope of the line bisecting the angle from line1, with slope 2, to line2, with no slope. Hope you'll help me. Thanks! Will wait asap.      Log On


   

Question 883651: Find the slope of the line bisecting the angle from line1, with slope 2, to line2, with no slope.
Hope you'll help me. Thanks! Will wait asap.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You are only given slopes, so it must mean that y-intercepts do not matter.
The lines could intersect each other anywhere, and they could intersect the x- and y-axes anywhere,
but for convenience I will make the lines intersect at the origin.
(Whoever does not like my drawing can move the axes up, down, and left or right to his/her taste. As long as nothing is rotated, the slopes remain the same).
The slope of line 1 is the tangent of the angle red%28A%29%29 it makes with the x-axis (measured counterclockwise from the positive x-axis).
The slope of the y-axis is undefined, and line 2 is the y-axis.
The bisector of the acute angle formed by lines 1 and 2 is drawn in green.
The angle it forms with the positive x-axis is
%28A%2B90%5Eo%29%2F2
(If you don't like degrees, it is %28A%2Bpi%2F2%29%2F2 in radians).
The slope of the bisector is the tangent of that angle.

If you only need an approximate value, you could use inverse trigonometric functions.
For an exact value, I would use trigonometric identities.

USING INVERSE FUNCTIONS:
We use the inverse of tangent from calculator or computer to find the angle that has a tangent of 2.
Hopefully you know how to do that with your calculator and/or computer.
The inverse function of tangent is represented as %22tan%22%5E%28-1%29 , or arctan , or ATAN , and in some calculators you have to press a key for inverse functions and then the tangent key.
tan%28A%29=2--->A=63.4%5Eo(rounded),
or A=1.107(rounded) in radians, if you prefer radians.
Using A=63.4%5Eo:
%28A%2B90%5Eo%29%2F2=76.7%5Eo and tan%2876.7%5Eo%29=4.23(rounded) is the approximate value for the slope of the bisector.
Using A=1.107(in radians):
%281.107%2Bpi%2F2%29%2F2=1.339(rounding the result but not pi ), and
tan%281.339%29=4.24(rounded) is the approximate value for the slope of the bisector.
You could round to more digits to get a better approximation.

USING TRIGONOMETRIC IDENTITIES FOR EXACT VALUE:
From the values of the trigonometric functions for A,
and the well known values for 90%5Eo=pi.2 ,
we would like to find the trigonometric functions for %28A%2B90%5Eo%29%2F2=%28A%2Bpi%2F2%29%2F2 .
For easier typing, I am going to use 90%5Eo+rather+%7B%7B%7Bpi%2F2 from now on.
There are probably many ways to get to tan%28%28A%2B90%5Eo%29%2F2%29 ,
but I am going to show the one that came to my mind first,
and that requires starting from sin%28A%29 and cos%28A%29 .
The right triangle with angle A that I included in my sketch
has legs measuring 1 and 2 ,
so the Pythagorean theorem says that the length of the hypotenuse is sqrt%281%5E2%2B2%5E2%29=sqrt%281%2B4%29=sqrt%285%29 .
Trigonometric ratios tell us that
sin%28A%29=2%2Fsqrt%285%29 and cos%28A%29=1%2Fsqrt%285%29
Trigonometric identities for sum of angles tell us that

We are half of he way there
Trigonometric identities for half angles are
abs%28sin%28B%2F2%29%29=sqrt%28%281-cos%28B%29%29%2F2%29 and abs%28cos%28B%2F2%29%29=sqrt%28%281%2Bcos%28B%29%29%2F2%29 ,
and we have to figure out the sign for ourselves.
Since %28A%2B90%5Eo%29%2F2 is in the first quadrant, we know that all its trigonometric functions have positive values, so


Tangent is sine divided by cosine, so
==sqrt%28%28sqrt%285%29%2B2%29%5E2%2F%28%28sqrt%285%29%2B2%29%28sqrt%285%29-2%29%29%29=sqrt%28%28sqrt%285%29%2B2%29%5E2%2F%285-2%5E2%29%29=sqrt%28%28sqrt%285%29%2B2%29%5E2%2F%285-4%29%29=sqrt%28%28sqrt%285%29%2B2%29%5E2%29=highlight%28sqrt%285%29%2B2%29
The approximate value of sqrt%285%29%2B2 is 4.236 .