# I'm stuck with some easy trigonometry

• August 2nd 2009, 02:06 AM
ibenot1337
I'm stuck with some easy trigonometry

If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg

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Find the largest angle of this triangle:
http://i431.photobucket.com/albums/q...ignmentq10.jpg

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Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.
• August 2nd 2009, 02:34 AM
e^(i*pi)
Quote:

Originally Posted by ibenot1337

If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg

---

Find the largest angle of this triangle:
http://i431.photobucket.com/albums/q...ignmentq10.jpg

---

Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.

$A = \frac{1}{2}acsin(B)$
• August 2nd 2009, 03:37 AM
jgv115
http://i431.photobucket.com/albums/q...ignmentq10.jpg
Let's change this up... A is B, B is C and C is A :)
Get the formula:
$a^2=b^2+c^2-2bc cosA$

transpose it since we want to find angles:

$a^2 + 2bcCosA=b^2+c^2$

$=2bcCosA=b^2+C^2-a^2$

$=CosA = \frac {b^2+c^2-a^2}{2bc}$

you with me?

alright time to substitute. You should know that the side opposite the angle will be the "corresponding" letter.

$CosA = \frac {9^2 + 15^2 - 10^2}{2*9*15}$

$CosA= \frac {206}{270}$

$CosA = 0.7629$

$A= Cos^-1 (0.7629)$

$A= 40'16'$

Now just do that for B one. Add A and B then subtract from 180. You have all your angles.
If anything is wrong here please correct me as I am in the stages of learning this too...
• August 2nd 2009, 03:42 AM
Failure
Quote:

Originally Posted by ibenot1337

If this triangle has an area of 24cm^2, what is the value of angle D? http://i431.photobucket.com/albums/q...signmentq8.jpg

You have already got a hint for this first problem from $\mathrm{e}^{\mathrm{i}\pi}$, so I won't elaborate on this one any further.

Quote:

Find the largest angle of this triangle:
http://i431.photobucket.com/albums/q...ignmentq10.jpg

Please show working in a simple way, I'm not too good at this and I really want to understand trigonometry.
Because the largest angle of a triangle is always opposite to the longest side, you are, in effect, asked to determine the angle at vertex A. Let me call this angle $\alpha$.
Now the law of cosine says that
$a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos(\alpha)$
where a, b, and c is the length of the side opposite to vertex A, B, and C, respectively.
Since all values in this equation, with the sole exception of $\alpha$, are given, you can simply solve for $\alpha$, like this:
$\alpha = \cos^{-1}\frac{b^2+c^2-a^2}{2\cdot b\cdot c}=\cos^{-1}\frac{10^2+9^2-15^2}{2\cdot 10\cdot 9}=\cos^{-1}\left(-\tfrac{11}{45}\right)\approx 104.15^\circ$
and are done!
• August 2nd 2009, 03:45 AM
jgv115
http://i431.photobucket.com/albums/q...signmentq8.jpg

For the first one use the formula above. I'll assume that B is D.

$A = \frac{1}{2}acsin(B)
$

transpose it to make SinB by itself:

$24=\frac {1}{2}acsin(B)$
$\frac {48}{ac} = SinB$

ok all done.

$\frac {48}{56} = SinB$

You can do the rest..
• August 28th 2009, 10:03 PM
ibenot1337
Thanks
Thank you for all your help.

I now understand the rules I need to solve those sorts of problems.(Hi)