# Obstuse trignometry

1. Jul 5, 2006

### vijay123

1) given that A is obstuse and that sinA=2/(sqaureroot of 5). find the value of cosA and tanA.
2)given that tanA=-5/12 and that tanA and cosA have opposite signs, find the values of sinA and of cosA.

CAN SOMEONE TEACH ME EXACTLY HOW TO DO THESE PROBLEMS WITHOUT USING TRGNOMETRIC IDENTITES.PLEASE....I AM NEW TO THIS TOPIC AND I WANT TO GET A HEADSTART OVER MY PEERS.
THANKS A LOT
VIJAY

2. Jul 5, 2006

### VietDao29

Ok, if A is obtuse, what sign will sin, cos, and tan of the angle A take (is it positive or negative?). You can take a look at the unit circle to see. Given sin A, one can manage finding cos A using Pythagorean Identity, i.e:
sin2 A + cos2 A = 1.
Having sin A, and cos A, can you find tan A? $$\tan A = \frac{\sin A}{\cos A}$$
I'll give you an example:
A is an acute angle. Given that sin A = 1 / 5. Find cos A, and tan A
Since A is an acite angle (i.e < 900), cos A must be positive. Using the Pythagorean Identity, we have:
$$\sin ^ 2 A + \cos ^ 2 A = 1$$
$$\Leftrightarrow \cos ^ 2 A = 1 - \sin ^ 2 A = 1 - \frac{1}{25} = \frac{24}{25}$$
$$\Rightarrow \cos A = \pm \frac{\sqrt{24}}{5}$$
Since cos A is positive, we choose:
$$\cos A = \frac{\sqrt{24}}{5}$$
We have:
$$\tan A = \frac{\sin A}{\cos A} = \frac{1}{\sqrt{24}}$$
Can you go from here? :)

3. Jul 5, 2006

### vijay123

Thanks A Lot For The Solution

4. Jul 5, 2006

### vijay123

vietdao...but is there any other way of doin such problems withou using the trignometric identities...because this problem is printed before the worksheet on trignometric identities....?

5. Jul 5, 2006

### vijay123

yes..thanks a lot...i can do it at last....but...does obtuse mean it is in the second quadrant?

6. Jul 5, 2006

### garyljc

Hello there ,
First thing in determining the signs , i suggest you a quadrant . Hence , there will be a first quadrant , second quandrant , third and fourth . Marking each quadrants , from 3 o'clock ( anticlockwise direction ) A , S , T , C (Where A=all postive , S=Only Sinus Postive , T=Only Tangent Positive , C=Only Cosinus Positive) .
Next you can then determine which quadrant the question is referring to . You can do without using trigonometric identities =) Try drawing and labelling the coordinates of your triangle , you should be able to find the answer =) Beware of the + and - signs .

7. Jul 5, 2006

### vijay123

ok...thanks a lot..

8. Jul 5, 2006

### HallsofIvy

Staff Emeritus
Draw a picture. Since you know $sin(x)= \frac{2}{\sqrt{5}}$ and sine is "opposite over hypotenuse", draw a right triangle with one side 2 and hypotenuse $\frac{1}{\sqrt{5}}$. You can use the Pythagorean theorem to find the third side and calculate all the other functions from that. Then use the fact that the angle is obtuse (and so in the second quadrant) to determine the sign.

9. Jul 6, 2006

### vijay123

yea...i get it....there r two ways of doing it..thanska lot