help on this pls

**watsonmath** · June 12th 2006, 03:43 AM

Actually i am just double checking my answer to see if i am doing it right. so anyone kindly help me out on these question. i just need the answer but working is prefered if you care to type it out

1. Find the angles of the triangle whose sides are of length 2, 4 and 5cm.
2.solve the triangle ABC in which A = 31degree , B = 14degree, and c = 7cm.
3.In triangle ABC , a = 4 , b = 6 , c = 9. Solve the triangle to find the missing elements.
4.In triangle ABC, B=122degree, C = 15degree, c= 2.5cm. Solve the triangle.

Simplify:
1)2sin^sqtx +cos2x
2)2sin2 0 cos2 0
3)(1-cos2A)/(sin2A)

2. Given sin x = -3/5 & cos y =-5/13;both X&Y are in 3rd quardrant. Find,without evaluating the angel.
a) sin(x+y)
b)cos(x-y)
c)tan(x+y)

solve the equation for all nonzero value of x for 0degree >x>360
3cos2x+sinx-2=0

Big thanks to you

**ThePerfectHacker** · June 12th 2006, 10:31 AM

Originally Posted by watsonmath

1. Find the angles of the triangle whose sides are of length 2, 4 and 5cm.

I am going to find one of the angles and leave the other two as excesirce (which I a nice way of saying I am too lazy to do the entire problem).

Okay, by the law of cosines you have that,
$5^2=4^2+2^2-2\cdot 2\cdot 4\cos x$ where $x$ is the angle formed between sides, 2 and 4.
Thus,
$25=16+4-16\cos x$
Thus,
$5=-16\cos x$
Thus,
$\cos x=-\frac{5}{16}$
Thus, using the inverse cosine function,
$x=\cos^{-1}(-5/16)\approx 108^o$

**ThePerfectHacker** · June 12th 2006, 10:39 AM

Originally Posted by watsonmath

Simplify:
1)2sin^sqtx +cos2x
2)2sin2 0 cos2 0
3)(1-cos2A)/(sin2A)

1)
Using double angle identity on cosine,
$2\sin^2x+\cos^2x-\sin^2x$
Adding,
$\sin^2x+\cos^2x=1$ by Pythagorean Identities.

2)Note that, $2\sin x\cos x=\sin 2x$ by double angle on sine.
Thus, $2\sin 2\theta \cos 2\theta=\sin 2(2\theta)=\sin 4\theta$

3)Split your fraction,
$\frac{1}{\sin 2A}-\frac{\cos 2A}{\sin 2A}=\csc 2A-\cot 2A$

**Soroban** · June 12th 2006, 12:12 PM

Hello, watsonmath!

2. Given $\sin x = -\frac{3}{5},\;\cos y = -\frac{5}{13}$ . . $x,y$ are in quadrant 3.

Find, without evaluating the angle: . $a)\;\sin(x+y)\qquad b)\;\cos(x-y)\qquad c)\;\tan(x+y)$

You're expected to know the compound angle formulas:
. . $\sin(A \pm B) \;=\;\sin(A)\cos(B) \pm \sin(B)\cos(A)$
. . $\cos(A \pm B) \;= \;\cos(A)\cos(B) \mp \sin(A)\sin(B)$

Since $\sin x = -\frac{3}{5}$ in quadrant 3, we have: $opp = -3,\;hyp = 5$
. . Hence: $adj = -4$ . . . and: $\cos x = -\frac{4}{5}$

Since $\cos y = -\frac{5}{13}$ in quadrant 3, we have: $adj = -3,\;hyp = 13$
. . Hence: $opp = -12$ . . . and $\sin y = -\frac{12}{13}$

$a)\;\sin(x + y)\;=\;\sin(x)\cos(y) + \sin(y)\cos(y) \;= \;\left(-\frac{3}{5}\right)$ $\left(-\frac{5}{13}\right) + \left(-\frac{12}{13}\right)\left(-\frac{4}{5}\right)$
. . . . . . . . . . . . $= \;\frac{15}{65} + \frac{48}{65}\;=\;-\frac{63}{65}$

$b)\;\cos(x + y)\;=\;\cos(x)\cos(y) - \sin(x)\sin(y)\;=\;\left(-\frac{4}{5}\right)$ $\left(-\frac{5}{13}\right) - \left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)$
. . . . . . . . . . . $= \;\frac{20}{65} - \frac{36}{65} \;= \;-\frac{16}{65}$

$c)\;\tan(x+y) \;= \;\frac{\sin(x+y)}{\cos(x+y)}\;= \;\frac{\frac{63}{65}}{\text{-}\frac{16}{65}} \;= \;-\frac{63}{16}$

**ThePerfectHacker** · June 12th 2006, 01:14 PM

Originally Posted by Soroban

Since $\sin x = -\frac{3}{5}$ in quadrant 3, we have: $opp = -3,\;hyp = 5$

I do not think it is proper to state that,
$\mbox{opp }=-3$ and $\mbox{hyp }=5$ but rather to state that the ordinate=-3 and radius vector=5.
Because as you know, trigonometric functions were redefined to coordinate definition rather than triangular definition to be able to work with positive and negative angles of any magnitude.

**Soroban** · June 12th 2006, 06:34 PM

Hello, ThePerfectHacker!

I do not think it is proper to state that: $\mbox{opp }=-3$ and $\mbox{hyp }=5$
but rather to state that the ordinate=-3 and radius vector=5. Because as you know,
trigonometric functions were redefined to coordinate definition rather than triangular definition
to be able to work with positive and negative angles of any magnitude.

I agree . . .

In my classes, I do define the trig functions in terms of $x,y,r$

A bit later, I remind them of the "old" language: opp-adj-hyp.
Once we determine the reference angle, I work with the "reference triangle"
. . (regardless of the quadrant), pointing out that these "sides" can be negative.
I've had very good results with this approach.

I hope that explains my rather casual terminology.

**ThePerfectHacker** · June 12th 2006, 07:00 PM

Originally Posted by Soroban

Hello, ThePerfectHacker!

I agree . . .

In my classes, I do define the trig functions in terms of $x,y,r$

A bit later, I remind them of the "old" language: opp-adj-hyp.
Once we determine the reference angle, I work with the "reference triangle"
. . (regardless of the quadrant), pointing out that these "sides" can be negative.
I've had very good results with this approach.

I hope that explains my rather casual terminology.

Do you think that way is more productive?
Because in your way it may get confusing to determine what is positive and what is negative and as you know high school students to not like to think, the standard terminology removes the tought process, which is what they want.

**watsonmath** · June 19th 2006, 02:52 AM

can you explain this question in detail a little(working). BTW.. the answer in my book is "tanA"
$<br /> \frac{1}{\sin 2A}-\frac{\cos 2A}{\sin 2A}=\csc 2A-\cot 2A<br />$

**Soroban** · June 19th 2006, 04:06 AM

Hello, watsonmath!

Simplify: $3)\;\frac{1-\cos2A}{\sin2A}$

The answer can be expressed in a variety of ways . . . all correct.
It's unfair for the book to have picked just one of them.

I'll show you how they got their answer . . .

They expected you to recognize these identities:

. . . $\sin^2\theta \;=\;\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad 1 - \cos2\theta \:=\:2\sin^2\theta$

. . . $\sin2\theta \;=\;2\sin\theta\cos\theta$

So that . $\frac{1 - \cos2A}{\sin2A}$ .becomes . $\frac{2\sin^2A}{2\sin A\cos A} \;= \;\frac{\sin A}{\cos A} \;= \;\tan A$

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