Here's the question you clicked on:
jazy
I'll put my question in the comments.
How would I find AB on here: |dw:1365267611167:dw|
If they give me these length measures. |dw:1365267723624:dw|
They're both right triangles. I have the lengths of the sides, but I want someone to explain how I could find AB, without giving me the answer.
sorry but the side lengths are not readable :/
or were they arrows ? :)
This is a matter of using Pythagoras' theorem. a^2 + b^2 = c^2 so name your sides, and then solve for all the things you know. Even if you don't know the side lengths, you can get a formula for the length of AB.
Yeah lol and bad ones too. Do you need the length measures to help me?
I would recommend you to use similar triangles here
I tried the Pythagoras theorem but neither one of my triangles have 2 side lengths given. :/
If there are no side lengths given then you need to find a formula. Instead of substituting the side length for the variable, just keep the variable in there.
Sorry, my brain is completely blank today when it comes to geometry. I'm not sure what you mean.
I need to clarify is CBA right angled?
Yes: |dw:1365268759222:dw|
Are u sure this is only the info given ?
These aren't the ones given to me but maybe they'll work: |dw:1365268927327:dw|
why not try this ? |dw:1365268967361:dw|
AB is the only unknown
Ok, I'll try it out on paper first. One sec..
I don't know what wrong with me today :/ So I get: \[\tan(A) = \frac{ 104 }{ x} = \frac{ 186 }{ x + 98 }\]But how could I solve this with two variables?? o.o
You only have one variable. A is irrelevant, all you want is x.
Oh, ok. So would I cross multiply now to solve for x?
Yes. Please give your medals to aravind :)
Ok and thanks(: \[\frac{ 104 }{ x } = \frac{ 186 }{ x + 98 } \to 104(x + 98) = 186x \to 104x + 192 = 186x \to 192 = 82x \to 2.34\]
what happened @jazy ?you seem terrible on math today :/
@AravindG I wish I knew. -.- Usually I'm pretty quick to catch on, but I guess today just isn't my day for math. I'll switch subjects for a while to refresh. Thanks for the patience and help, I really do appreciate it. c: