Question

Find the lengths UQ, VR, and TP on triangle SWN.

Answer 1

@trollz1337 is the answer to be a number or an equation?

Answer 2

number

Answer 3

is there any other information given?

Answer 4

and is this in a geometry class or linear algebra?

Answer 5

geomtry should i be somewhere else how do i change it

Answer 6

you are in the right spot, i asked to get an idea of what methods to use ^_^

Answer 7

ya i was asking what method to use
to solve it also

Answer 8

did u look at the picture?

Answer 9

do you know how to solve it?

Answer 10

i'm trying to figure it out now...

Answer 11

ok

Answer 12

suggestions?
@agent0smith @terenzreignz @AllTehMaffs

Answer 13

It has do with the ratios of the sides to the base. (bad picture to draw on) They're all similar triangles, so they should all scale with the sides. I just can't remember right quick what to do. 1 sec...

Answer 14

@texaschic101

Answer 15

use law of cosines

\[ 26^2 = (4n)^2+(4m)^2-2(4n)(4m)cosK\]

K is the angle WSN in original pic

Set cosines equal to each other for respective triangles

Answer 16

this question has got to be a joke!

Answer 17

It's obnoxious

So here;

\[ 26^2 = 16n^2 + 16m^2 - 32nm \cos K
\\ \
\\ \frac{26^2 - 16n^2-16m^2}{32} = -nm\cos K\]
Then to find C (they call it VR)
\[ C^2 = n^2 + m^2 - 2nm \cos K
\\ \
\\ \frac{C^2 - n^2-m^2}{2} = -nm\cos K\]
Set them equal to each other

\[ \frac{26^2 - 16n^2-16m^2}{32} = \frac{C^2 - n^2-m^2}{2}\]

The n's and m's cancel out, and you're left with
\[ C^2 = \frac{26^2}{16} = \left( \frac{26}{4} \right) ^2 \]

Answer 18

So it scales directly with the sides.

Answer 19

Which might just be a property of similar triangles..... :/

Answer 20

\[a) \ \ \ UQ = 13
\\ \
\\ b) \ \ \ VR = 13/2
\\ \
\\ c) \ \ \ TP = 26/3 \]

Answer 21

@alltehmaffs wow, i'm impressed!

Answer 22

:) thanks

Answer 23

@alltehmaffs
i'm logging off, gotta eat dinner, thanks for looking at this!

Answer 24

I think it should be. I'm trying to make sure though

Answer 25

Good eats at ya. I'll see what I can do :)

Answer 26

@trollz1337 This all make sense to you? have you used the law of cosines before?

Answer 27

oh, the kid left.
bye bye

Answer 28

I solved your pic real quick

Answer 29

2n
\[ 2n \sin \theta = a
\\ 4n \sin \theta = b
\\ \
\\2 (2 \sin \theta) = 2 (b/2)
\\ 2 \sin \theta = b/2 = a
\\ b = 2a \]