I completing complex math via correspondence .. don't have a teacher to refer to, so if you could check me equation/answer and advise would be much appreciated.

Printable View

- Jun 17th 2008, 04:38 PMRisrocksCan you check my answer Please.
I completing complex math via correspondence .. don't have a teacher to refer to, so if you could check me equation/answer and advise would be much appreciated.

- Jun 17th 2008, 11:24 PMReckoner
I honestly can't make heads or tails of your work. It isn't clear to me at all what you are doing. Where did $\displaystyle ab = 2\sqrt{10^2 + 10^2}$ come from? And you can't use the law of cosines that way--you need 3 sides of the

*same*triangle.

Anyway, here is how I would do it:

First, I suggest finding the height of the "roof" (that is, the length of the perpendicular between $\displaystyle a$ and $\displaystyle b$). Call it $\displaystyle h$, and we have:

$\displaystyle \tan22^\circ = \frac h{10000+15000}$

$\displaystyle \Rightarrow h = 25000\tan22^\circ\approx10100.66$

Now everything follows from the Pythagorean theorem:

$\displaystyle a^2 = 10000^2 + 25000^2\tan^222^\circ$

$\displaystyle \Rightarrow a = 5000\sqrt{25\tan^222^\circ+4}\approx14213.488$

$\displaystyle c$ is found in much the same way, and $\displaystyle b = a$ since the two inner right triangles are congruent (two sides and an included angle equal). - Jun 18th 2008, 01:03 AMRisrocks
- Jun 18th 2008, 01:06 AMjanvdl
- Jun 18th 2008, 02:23 AMRisrocks
- Jun 18th 2008, 10:38 AMReckoner
I never meant to discourage you! I only wanted to nudge you in the right direction. Keep in mind that there are plenty of people that have difficulty grasping basic algebra, or even simple arithmetic, and you seem to at least know what you are doing there if you are doing trigonometry. Even the greatest mathematicians face difficulties, and yours can be overcome as long as you put in the necessary effort.

Now, I wouldn't expect you to "see" how I made that jump because I skipped a bunch of steps in-between. But the idea was that you could get to the same answer if you used the same starting point. Here is what I did:

$\displaystyle a^2 = 10000^2 + 25000^2\tan^222^\circ$

$\displaystyle \Rightarrow a^2 = 100000000 + 625000000\tan^222^\circ$ (Evaluating exponents)

$\displaystyle \Rightarrow a = \sqrt{100000000 + 625000000\tan^222^\circ}$ (Taking square root of both sides; we can ignore the negative root since $\displaystyle a > 0$)

$\displaystyle \Rightarrow a = \sqrt{4\cdot25000000 + 25\cdot25000000\tan^222^\circ}$ (Factoring)

$\displaystyle \Rightarrow a = \sqrt{25000000\left(4 + 25\tan^222^\circ\right)}$ (Factoring)

$\displaystyle \Rightarrow a = \sqrt{25000000}\sqrt{4 + 25\tan^222^\circ}$ (Splitting square root: $\displaystyle \sqrt{ab}=\sqrt a\sqrt b$)

$\displaystyle \Rightarrow a = 5000\sqrt{4 + 25\tan^222^\circ}$ (Evaluating: $\displaystyle \sqrt{25000000} = \sqrt{25\cdot10^6} = \sqrt{25}\sqrt{10^6} = 5\cdot10^3 = 5000$)

And of course, if you are just looking for an approximation, you could have stopped at the first step and threw it into the calculator. I just like keeping things in their exact value.

- Jun 18th 2008, 10:46 AMmasters
Edit: Just trying to add a little flavor, Reckoner. Too slow, tho.

From Reckoner's last post, he found the altitude h.

$\displaystyle h = 25000\tan22$

One side of the right triangle is already known to be 10000, and now we can find the hypotenuse "a" using the Pathagorean theorem.

$\displaystyle a^2=(10000)^2 + h^2$

Substituting for h, we have:

$\displaystyle a^2=(10000)^2+(25000\tan22)^2$

$\displaystyle a=\sqrt{(10000)^2+(25000\tan22)^2}$ - Jun 18th 2008, 05:45 PMRisrocks
- Jun 18th 2008, 05:56 PMRisrocks