Help me solve a trig problem

[lpaine1331]

ok, its been along time. I n eed a little help getting started with this trig problem.
I have solved the problem all the way to this equation and I have to now solve for theta (the angle).

1.226 tan^2 theta + 24 tan theta - .575 = 0

How do i do this again?

thanks alot

 

[khaila]

Re: trig question

1.226 tan^2 theta + 24 tan theta - .575 = 0

How do i do this again?

thanks alot

The soluation seems to be:
to bring the upon eqution into the form

aX^2+bX+c=0
"a" is can be removed bymultiplying it with "c" and at the final solution you will divide the result agian with "a"


while
X^2=tan^2 theta
a=1.226 -->
b=24
c=.575

continue...

 

[tny21]

Re: trig question

don't know if this helps but I plugged this in to my TI-89:

solve(1.226*tan(x))^2+24*tan(x)-.575=0,x)

the calculation resulted with:
x=92.9208 or x=1.37077 or x=-87.0782 degrees

 

[aersoy]

Re: trig question

tan^2(θ)+24tan(θ)-0.575=0
>>>>
tan(θ)=s
>>>> s^2+24s-0.575=0

>> s ≈- 0. 024, s≈-19.552
by using arctan == inverse tan
tan(θ) = -0.024 >>>> θ = arctan (-0.024) ≈-1.37°
tan(θ) = -19.552 >>> θ = arctan (-19.552) ≈-87.07°

 

[hamidrezakarami]

Re: trig question

aersoy's solution is true.

 

[srieda]

Re: trig question

ok, its been along time. I n eed a little help getting started with this trig problem.
I have solved the problem all the way to this equation and I have to now solve for theta (the angle).

1.226 tan^2 theta + 24 tan theta - .575 = 0

How do i do this again?

thanks alot

let tan Θ = x...
then 1.226 x² + 24 x - 0.575 = 0

x = (-24 ± √(24² - 4×1.226×(-0.575))) / (2 × 1.226)
= (-24 ± 24.058) / 2.452
x = 0.0237 or -19.6

Θ = 1.3576° or -87°

 

[jonw0224]

Re: trig question

Since the tangent function repeats every 180 degrees, the full solution is:

tan x = 0.0237
x = 1.36° + 180° * n
where n is any integer

and

tan x = -19.6
x = -87.1° + 180° * n
where n is any integer

That explains the 92°.

-jonathan

 

[pareshatlnmiit]

Re: trig question

Sreida is right ,always remember whenever you met a situation like that where you can make a quadratic or cubic then go for that ,you will always get a derired output

 

[MSRA]

Re: trig question

Its Right i think That's the only way.....

tan^2(θ)+24tan(θ)-0.575=0
>>>>
tan(θ)=s
>>>> s^2+24s-0.575=0

>> s ≈- 0. 024, s≈-19.552
by using arctan == inverse tan
tan(θ) = -0.024 >>>> θ = arctan (-0.024) ≈-1.37°
tan(θ) = -19.552 >>> θ = arctan (-19.552) ≈-87.07°