trig, thx (1 Viewer)

[tommykins]

回复: Re: trig, thx

ok... thanks that sorta helps

um can you help me with this question

question:
tan 50degrees

answer:
tan 150degrees= tan (180-30)
acute angle=30degrees
tan is negative in quadrant 2

(I dont get how it is in quad 2, thats the only prob)

can someone plz help out here?

what's the question asking exactly...

 

[kingnothing999]

hmm haha, i think i get it now

um, but do you know how to do this question:

If sin degrees= 4/7 and tan degrees < 0, find the exact value of cos degrees and tan degrees

sorry about the degrees, i duno how to type the degrees thing

 

[tommykins]

回复: Re: trig, thx

hmm haha, i think i get it now

um, but do you know how to do this question:

If sin degrees= 4/7 and tan degrees < 0, find the exact value of cos degrees and tan degrees

sorry about the degrees, i duno how to type the degrees thing

Draw up a triangle with the opposite angle 4 and the hypotenuse 7.

the adjacent side becomes sqrt(49-16) = sqrt(33) [using pythag's]

since tan x < 0 but sin x > 0, we are in the 2nd quadrant.

cosx < 0 in 2nd quad, so cos x = -sqrt33/7
tanx < 0 so tanx = -4/sqrt33

 

[kingnothing999]

wow, that really helped me, thanks alot

 

[kingnothing999]

hey, does any1 know how to do this question, its from
trigonometric equations:

Q1 Solve for -180degrees < theatre <180degrees


f) 3 tan square root 2 theatre = 1


hope you understand what im askin, if not just tell me

 

[kingnothing999]

Do you mean 3 tan (square root (2) @) = 1 (-180^o < @ < 180^o)?

If so, tan (square root (2) @) = 1/3 [DIVIDE EACH SIDE BY 3]
-180 root2^o < root2@ < 180 root2^o [CHANGE DOMAIN TO SUIT WHAT'S INSIDE THE BRACKETS I.E. MULTIPLY EACH PART OF DOMAIN EQUATION BY ROOT2]

root2 @ = tan^-1 (1/3) [TAKE THE TAN^-1 OF EACH SIDE]
= 18^o26'5.82'' [CALCULATOR DISPLAY]
= 18^o26' [ROUNDED TO NEAREST MINUTE]

WORKING FROM QUADRANTS:

root2 @ = 18^o26', 198^o26', -161^o34' <--- 18 IS SMALL ANGLE, SINCE DOMAIN IS ROOT2 x 180, 18 + 180 FITS DOMAINS AND 18 - 180 ALSO FITS DOMAIN
-180 < @ < 180 [CHANGING DOMAIN BACK TO ORIGINAL ONE AS EXPRESSED IN THE QUESTION]
@ = 13^o2'3.61'', 140^o18'48.8'', -114^o14'41.59'' [DIVIDE EACH ANSWER ABOVE BY ROOT 2 TO GAIN VALUES FOR @]
@ = 13^o2', 140^o19', -114^o15' [ROUNDED ANSWERS TO NEAREST MINUTE]

If you have time check your answers i.e. substitute your values for @ back into the original equation. Since you have rounded your answer won't be exactly on the dot but it should be really close. In this case when you substitute @ in the answers you get are 0.9998, 1.0002 and 0.9995 which are virtually 1 [right hand side of equation]. So you can be 99.99% sure these values for @ are correct.

Have I missed any answers?

hmm, well.. the answer is like:
@ or theatre= + or - 30 or 150<SUP>o </SUP>
<SUP></SUP>
<SUP></SUP>
Its got to do with angle of magnitude, and the ASTC rule
( All station to central) and also exact ratios such as
sin30<SUP>o</SUP>= 1/2
tan 45<SUP>o</SUP>= 1
etc..

Il try to write the question clearer to you:

Solve for -180<SUP>o</SUP> < @ < 180<SUP>o</SUP>
<SUP></SUP>
3 tan<SUP>2 </SUP>@ = 1 (this is exaclty the way the question is set out in the textbook)
<SUP></SUP>
<SUP></SUP>
<SUP>hope this is clearer to you</SUP>

 

[bored of sc]

3 tan<SUP>2 </SUP>@ = 1 {-180<SUP>o</SUP> < @ < 180<SUP>o</SUP>}
<SUP>tan² @ = 1/3 [divide each side by 3]</SUP>
<SUP>tan @ = + 1/root3 [take square root of each side]</SUP>
<SUP>@ = tan^-1 (1/root3)</SUP> [take tan^-1 of each side]
<SUP>= 30^o</SUP>
<SUP></SUP>
<SUP>Now to the different magnitudes. </SUP>
<SUP>Since tan @ = + 1/root3 the possible answers are...</SUP>
<SUP>= 30^o, 0-30^o, 180-30^o, 0-(180-30)^o</SUP>
<SUP>@ = + 30^o, + 150^o</SUP>

Explanation of answers:
30 is the original, small angle that was gained from the calculator.
Since @ = tan^-1 + (1/root3) you have to take the 30^o angle from every quadrant. Since the domain is {-180<SUP>o</SUP> < @ < 180<SUP>o</SUP>} you will be using the 1st and 2nd quadrants moving in an anticlockwise, positive direction while for the other two answers you will be using the 4th and 3rd quadrants moving in an clockwise, negative direction. You work from the x-axis where the angle is 0 or 180. So to gain answers you do 30 (first quadrant), 180-30 (second quadrant), 0-30 (fourth quadrant), 0-(180-30) (third quadrant).
Hope that helps.

 

[kingnothing999]

thanks, that really helped