Why Use Logarithms to Solve Trig Problems?

[mishek]

Hi,

Can somebody please point me into a direction how to solve the attached trig problem?

First step, where can I find such values for sin (alpha)?Thanks.

 

[Mentallic]

Hi,

Can somebody please point me into a direction how to solve the attached trig problem?

First step, where can I find such values for sin (alpha)?Thanks.

You'd use a calculator to find the value of [itex]\sin(\alpha)[/itex]. The rest of the information can be inferred by what you've already been given (and with the use of a calculator). For example, you can find a since you know c and [itex]\alpha[/itex], you can also find [itex]\beta[/itex] since you know [itex]\alpha[/itex], and hence you can finally find b since you know c and [itex]\beta[/itex].

 

[mishek]

Hi Mentallic,

Thanks for a quick reply.

I know I can use a calculator :), but I would like to know how to solve it in a manner that is given on a photo.

There must be some tables that I should use (i assume), but i don't know which one.

Also, the value of c in step II, where did that come from?

 

[Samy_A]

Hi Mentallic,

Thanks for a quick reply.

I know I can use a calculator :), but I would like to know how to solve it in a manner that is given on a photo.

There must be some tables that I should use (i assume), but i don't know which one.

Also, the value of c in step II, where did that come from?

On the right side they give the logarithms (to base 10) of the values.
For example ##\log_{10}(c)=\log_{10}(457)=2.65992...##

 

[Mentallic]

Hi Mentallic,

Thanks for a quick reply.

I know I can use a calculator :), but I would like to know how to solve it in a manner that is given on a photo.

There must be some tables that I should use (i assume), but i don't know which one.

It depends on how much accuracy you want. You can just google search "sine table" and you'll find plenty of tables that offer each degree from 0 to 90, but assuming you want more accuracy since you're given angles that involves minutes and seconds too, in the case of [itex]\alpha[/itex] personally, I'd be satisfied with approximating [itex]\alpha = 32^o40'15''[/itex] into [itex]\alpha \approx 32\frac{2}{3}^o[/itex] and just calculate [itex]\frac{1}{3}\sin(32^o)+\frac{2}{3}\sin(33^o)[/itex]. If this doesn't make sense for you, then have a look at linear interpolation: https://en.wikipedia.org/wiki/Mathematical_table

Also, the value of c in step II, where did that come from?

I have no idea. The values in those boxes don't coincide with what the given values of c and alpha at the top were. I wouldn't dwell too heavily on it though since you can quite easily re-do that entire page for yourself with the correct values.

edit: Good catch

On the right side they give the logarithms (to base 10) of the values.
For example ##\log_{10}(c)=\log_{10}(457)=2.65992...##

 

[mishek]

On the right side they give the logarithms (to base 10) of the values.
For example ##\log_{10}(c)=\log_{10}(457)=2.65992...##

Hi Samy_A, big thanks for the reply!

When I catch some time, i'll go in that direction and try to solve it myself.

But i am wondering, why use logarithms?

 

[Samy_A]

Hi Samy_A, big thanks for the reply!

When I catch some time, i'll go in that direction and try to solve it myself.

But i am wondering, why use logarithms?

That's how we solved this kind of problems when I was in high school many many years ago, just before calculators became ubiquitous.
We had tables that gave logarithms, the trigonometric values, their logarithms, ...
We used them to look up the values, and logarithms were used to convert multiplications into additions.

No idea why one would do it this way now.