Trigonometry and geometry assignment

[niketriantafillidis@gmail]

Hi! Would appreciate some help on this. Don't know where to start without any given angles.

 

[Deleted member 4993]

Hi! Would appreciate some help on this. Don't know where to start without any given angles.

View attachment 32312

Can you post an image of the exact assignment ? I suspect some information is missing. As posted the rectangle is not unique.

 

[niketriantafillidis@gmail]

I had this assignment on a resent test, which I don't have access to and they only gave us the answers to the questions and not the calculations. I'm not a 100% sure but there might have been a 90 degree angle in the bottom right corner and maybe one in the left one at the top. The answer is suppose to på A=B=10.

 

[LCKurtz]

You can get A and B = 10 if you assume the lower right corner is a right angle. Look at the diagram below:

Since ABC is a right triangle with legs of [imath]\sqrt {10}[/imath] you can calculate its hypotenuse to be [imath]\sqrt{20}[/imath]. You are given sides AD and CD so you can calculate [imath]\cos a[/imath] using the law of cosines. Then use the addition formula for [imath]\cos V = \cos(a + \frac \pi 4)[/imath] and simplify it to get your answer.

 

[The Highlander]

I had this assignment on a resent test, which I don't have access to and they only gave us the answers to the questions and not the calculations. I'm not a 100% sure but there might have been a 90 degree angle in the bottom right corner and maybe one in the left one at the top. The answer is suppose to på A=B=10.

You can get A and B = 10 if you assume the lower right corner is a right angle. Look at the diagram below:

Since ABC is a right triangle with legs of [imath]\sqrt {10}[/imath] you can calculate its hypotenuse to be [imath]\sqrt{20}[/imath]. You are given sides AD and CD so you can calculate [imath]\cos a[/imath] using the law of cosines. Then use the addition formula for [imath]\cos V = \cos(a + \frac \pi 4)[/imath] and simplify it to get your answer.

What are A & B, indeed?​

You don’t mention A & B anywhere on your sketch! ?

So how are we supposed to know what is to be calculated? ?

Assuming there is a right angle in the bottom right corner then there cannot also be one in the top left or the side lengths wouldn’t be consistent with the those provided. Furthermore, if the answers given (A=B=10 & cosV = \(\displaystyle \frac{\sqrt{A}}{B}\)) are correct then cos V = \(\displaystyle \frac{\sqrt{10}}{10}\Rightarrow\) V ≈ 71.6°. If V were obtuse (as per LCK's sketch) then cos V would be negative, so I would suggest the figure should look more like this:-

​

Unlike LCK, I would name the vertices in a clockwise direction (as per the second figure) but that’s nothing more than a matter of preference, however, given that my sketch appears to fit the dimensions (& answers) provided, so far, I am still at a loss to know what the A & B actually are!

Are A & B line lengths or angles or areas or what? ​

 

[Dr.Peterson]

What are A & B, indeed?​

You don’t mention A & B anywhere on your sketch! ?

You're misreading the problem. It doesn't say A and B are angles or lengths. It says this:



A and B are the radicand and the denominator in the answer, written in this form; since the answer (under @LCKurtz's interpretation) is [math]\cos(v)=\frac{\sqrt{10}}{10}[/math]
we get [imath]A=10, B=10[/imath].

 

[niketriantafillidis@gmail]

You can get A and B = 10 if you assume the lower right corner is a right angle. Look at the diagram below:
View attachment 32321
Since ABC is a right triangle with legs of [imath]\sqrt {10}[/imath] you can calculate its hypotenuse to be [imath]\sqrt{20}[/imath]. You are given sides AD and CD so you can calculate [imath]\cos a[/imath] using the law of cosines. Then use the addition formula for [imath]\cos V = \cos(a + \frac \pi 4)[/imath] and simplify it to get your answer.

What are A & B, indeed?​

You don’t mention A & B anywhere on your sketch! ?

So how are we supposed to know what is to be calculated? ?

Assuming there is a right angle in the bottom right corner then there cannot also be one in the top left or the side lengths wouldn’t be consistent with the those provided. Furthermore, if the answers given (A=B=10 & cosV = \(\displaystyle \frac{\sqrt{A}}{B}\)) are correct then cos V = \(\displaystyle \frac{\sqrt{10}}{10}\Rightarrow\) V ≈ 71.6°. If V were obtuse (as per LCK's sketch) then cos V would be negative, so I would suggest the figure should look more like this:-

View attachment 32335View attachment 32331​

Unlike LCK, I would name the vertices in a clockwise direction (as per the second figure) but that’s nothing more than a matter of preference, however, given that my sketch appears to fit the dimensions (& answers) provided, so far, I am still at a loss to know what the A & B actually are!

Are A & B line lengths or angles or areas or what?
​

PS: I have no idea why that "View attachment 32331" appears above! ?

Thanks for your answers!
Like I said, I don't have access to the test and I don't remember exactly how the question was written or what information was given, which makes this a bit of a guessing game (my apologies for that one). But like you both have written, the bottom right corner must have been a right angle, otherwise I'm not sure how I'm suppose to solve it.
Would also like to add that the answer A=10 B=10 only was an example of what the answer could be, don't know if that changes anything though.
Either way, thanks for your help! I managed to get the right answer with what you both have written.

 

[LCKurtz]

What are A & B, indeed?​

You don’t mention A & B anywhere on your sketch! ?

So how are we supposed to know what is to be calculated? ?

Assuming there is a right angle in the bottom right corner then there cannot also be one in the top left or the side lengths wouldn’t be consistent with the those provided. Furthermore, if the answers given (A=B=10 & cosV = \(\displaystyle \frac{\sqrt{A}}{B}\)) are correct then cos V = \(\displaystyle \frac{\sqrt{10}}{10}\Rightarrow\) V ≈ 71.6°. If V were obtuse (as per LCK's sketch) then cos V would be negative, so I would suggest the figure should look more like this:-

View attachment 32335View attachment 32331​

Unlike LCK, I would name the vertices in a clockwise direction (as per the second figure) but that’s nothing more than a matter of preference, however, given that my sketch appears to fit the dimensions (& answers) provided, so far, I am still at a loss to know what the A & B actually are!

Are A & B line lengths or angles or areas or what?
​

PS: I have no idea why that "View attachment 32331" appears above! ?

Since you seem to have observed I named the vertices in a counterclockwise direction ABCD and I mention triangle ABC, I think you know they are labels of the vertices. When I drew the figure in another application I forgot the OP had used A and B to represent numbers in his answer. Mild slip-up. Please try not to hyperventilate too much about it.

 

[The Highlander]

You're misreading the problem. It doesn't say A and B are angles or lengths. It says this:

A and B are the radicand and the denominator in the answer, written in this form; since the answer (under @LCKurtz's interpretation) is [math]\cos(v)=\frac{\sqrt{10}}{10}[/math]
we get [imath]A=10, B=10[/imath].

Since you seem to have observed I named the vertices in a counterclockwise direction ABCD and I mention triangle ABC, I think you know they are labels of the vertices. When I drew the figure in another application I forgot the OP had used A and B to represent numbers in his answer. Mild slip-up. Please try not to hyperventilate too much about it.

You’re quite right, of course, Dr.P.

I completely failed to notice the “if” in the second line! ?

Most likely an age & eyesight problem (that’s why I use a larger font in my posts; so I can read ‘em) but, nevertheless, mea culpa! ?

Apologies too to LCK. I certainly didn’t intend to cause any offence regarding your naming of the vertices (I did say it was merely a matter of “preference”). It was your sketch (and the OP’s) that first caught my attention (another possible reason why I missed that all important “if”? My own “Mild slip-up”.)

I was concentrating more on the sketch(es) because, given the dimensions provided, I couldn’t see how V might be obtuse; unless anyone can show me how to construct a quadrilateral (not a “rectangle”, as Subhotosh suggested ?) with the side lengths provided that has an obtuse angle at Point A (and a right angle in the corner we have (all?) now assumed)? ?

That was what first prompted me to give any thought to the problem at all. ?

I wasn’t particularly “hyperventilating” about that either, it just struck me as an obvious anomaly. ?

And, of course, I agree entirely with your (LCK’s) solution to the problem (using your nomenclature ?):-

By the Cosine Rule: \(\displaystyle cos a=\frac{AD^2+AC^2-DC^2}{2×AD×AC}=\frac{16+20-4}{2×4×\sqrt{20}}=\frac{32}{8\sqrt{20}}=\frac{4\sqrt{20}}{20} \Rightarrow cos a=\frac{\sqrt{20}}{5} \\ \Rightarrow sin a=\frac{\sqrt{5}}{5}   \left(\frac{5}{25}+\frac{20}{25}=1\right)\)

And, since V= (a+45°), then:-

\(\displaystyle cos V=cos (a+45°)=cos a×cos 45°-sin a×sin 45° \\      =\frac{\sqrt{20}}{5}×\frac{1}{\sqrt{2}}-\frac{\sqrt{5}}{5}×\frac{1}{\sqrt{2}}=\frac{\sqrt{20}-\sqrt{5}}{5\sqrt{2}}=\frac{2\sqrt{5}-\sqrt{5}}{5\sqrt{2}}=\frac{\sqrt{5}}{5\sqrt{2}}=\frac{\sqrt{5}\sqrt{2}}{10}=\frac{\sqrt{10}}{10}\)

Hence, A=B=10. (Q.E.D) ?

I trust we are all now in agreement and no (further) umbrage taken? ?

Friends, huh? ?