# Can you check my answer Please.

• June 17th 2008, 04:38 PM
Risrocks
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.
• June 17th 2008, 11:24 PM
Reckoner
Quote:

Originally Posted by Risrocks
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.

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 $ab = 2\sqrt{10^2 + 10^2}$ 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 $a$ and $b$). Call it $h$, and we have:

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

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

Now everything follows from the Pythagorean theorem:

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

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

$c$ is found in much the same way, and $b = a$ since the two inner right triangles are congruent (two sides and an included angle equal).
• June 18th 2008, 01:03 AM
Risrocks
Quote:

Originally Posted by Reckoner
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 $ab = 2\sqrt{10^2 + 10^2}$ 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 $a$ and $b$). Call it $h$, and we have:

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

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

Now everything follows from the Pythagorean theorem:

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

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

$c$ is found in much the same way, and $b = a$ since the two inner right triangles are congruent (two sides and an included angle equal).

Sorry .... I think I should just give up on this whole complex math. The info you have provided me is great ... I'm really trying my best, but obviously my brain is not equipped to deal with this. Thanks again.
• June 18th 2008, 01:06 AM
janvdl
Quote:

Originally Posted by Risrocks
Sorry .... I think I should just give up on this whole complex math. The info you have provided me is great ... I'm really trying my best, but obviously my brain is not equipped to deal with this. Thanks again.

The last thing you should dare consider is giving up. Keep working on it. A little bit more practise and you'll get it soon.
• June 18th 2008, 02:23 AM
Risrocks
Quote:

Originally Posted by Reckoner
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 $ab = 2\sqrt{10^2 + 10^2}$ 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 $a$ and $b$). Call it $h$, and we have:

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

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

Now everything follows from the Pythagorean theorem:

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

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

$c$ is found in much the same way, and $b = a$ since the two inner right triangles are congruent (two sides and an included angle equal).

By the way ... this is how frustrating this is for me (and you having to answer me)! I can't grasp how from:
$a^2 = 10000^2 + 25000^2\tan^222^\circ$
you get the
5000\sqrt etc... etc...
I should just give up right?
• June 18th 2008, 10:38 AM
Reckoner
Quote:

Originally Posted by Risrocks
By the way ... this is how frustrating this is for me (and you having to answer me)! I can't grasp how from:
$a^2 = 10000^2 + 25000^2\tan^222^\circ$
you get the
5000\sqrt etc... etc...
I should just give up right?

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:

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

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

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

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

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

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

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

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.
• June 18th 2008, 10:46 AM
masters
Quote:

Originally Posted by Risrocks
By the way ... this is how frustrating this is for me (and you having to answer me)! I can't grasp how from:
$a^2 = 10000^2 + 25000^2\tan^222^\circ$
you get the
5000\sqrt etc... etc...
I should just give up right?

Edit: Just trying to add a little flavor, Reckoner. Too slow, tho.

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

$h = 25000\tan22$

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

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

Substituting for h, we have:

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

$a=\sqrt{(10000)^2+(25000\tan22)^2}$
• June 18th 2008, 05:45 PM
Risrocks
Quote:

Originally Posted by janvdl
The last thing you should dare consider is giving up. Keep working on it. A little bit more practise and you'll get it soon.

Thanks for your words of encouragement ... it was late last night when I posted my comment ... new day today with a new outlook to attempt and conquer my studies.
• June 18th 2008, 05:56 PM
Risrocks
Quote:

Originally Posted by Reckoner
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:

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

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

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

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

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

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

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

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.

My apologies ... you in no way have discouraged me - it was late last night when I was studying and tiredness & frustration got the better of me. I will analyse your latest reply, and again thank you for taking the time to assist.