Linear Approximation and Triangles

[sarahjohnson]

http://postimg.org/image/kph8hls05/

I do not understand what the problem is asking. For the first part, how did they get the square roots?

Can you do all parts of the problem please?

Thank you for your help.

 

[HallsofIvy]

The problem says "At first the angle A is \(\displaystyle \pi/4\) radians = 45 degrees and B and the angle B is \(\displaystyle \pi/3\) radians= 60 degrees." They expect you to know that \(\displaystyle sin(\pi/4)= \sqrt{2}/2\) and \(\displaystyle sin(\pi/3)= \sqrt{3}/2\). Dividing, \(\displaystyle \frac{sin(A)}{sin(B)}= \frac{\sqrt{3}}{2}\frac{2}{\sqrt{2}}= \frac{\sqrt{3}}{\sqrt{2}}\).

 

[sarahjohnson]

Ok that makes sense, thank you

but what do I put into the boxes for implicit differentiation if sinA/sinB= that...

I'm sorry if this is a dumb question... I still don't understand what I am suppose to take the derivative of

 

[Deleted member 4993]

Ok that makes sense, thank you

but what do I put into the boxes for implicit differentiation if sinA/sinB= that...

I'm sorry if this is a dumb question... I still don't understand what I am suppose to take the derivative of

sin(B) = (b/a) * sin(A) .........................................................................................(1)

B = sin^-1[(b/a) * sin(A)]

Now differentiate B w.r.t. A to find dB/dA ..................... or do implicit differentiation of (1)

The question is very clear - in my opinion cannot be expressed any "clearer".

 

[lookagain]

They expect you to know that \(\displaystyle sin(\pi/4)= \sqrt{2}\)/\(\displaystyle \ 2\) and \(\displaystyle sin(\pi/3)= \sqrt{3}/2\). Dividing, \(\displaystyle \frac{sin(A)}{sin(B)}= \frac{\sqrt{3}}{2}\cdot\frac{2}{\sqrt{2}}= \frac{\sqrt{3}}{\sqrt{2}}\)

.
(Originally missing "/" for a fraction bar and missing some symbol to indicate multiplication between two fractions.)

 

[Deleted member 4993]

The problem says "At first the angle A is \(\displaystyle \pi/4\) radians = 45 degrees and B and the angle B is \(\displaystyle \pi/3\) radians= 60 degrees." They expect you to know that \(\displaystyle sin(\pi/4)= \sqrt{2}/2\) = sin(A)and \(\displaystyle sin(\pi/3)= \sqrt{3}/2\) = = sin(B). Dividing, \(\displaystyle \frac{sin(A)}{sin(B)}= \frac{\sqrt{3}}{2}\frac{2}{\sqrt{2}}= \frac{\sqrt{3}}{\sqrt{2}}\).

A small correction:

\(\displaystyle \frac{sin(A)}{sin(B)}= \ \dfrac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} \ = \frac{\sqrt{2}}{\sqrt{3}}\)

\(\displaystyle \dfrac{sin(B)}{sin(A)} \ = \ \dfrac{\sqrt{3}}{\sqrt{2}}\)

\(\displaystyle sin(B) \ = \ \dfrac{\sqrt{3}}{\sqrt{2}} \ * \ sin(A)\)

\(\displaystyle \cos(B) \ \dfrac{dB}{dA} \ = \ \dfrac{\sqrt{3}}{\sqrt{2}} \ * \cos(A)\)

Now continue.....