In triangle TVW below, what is the length of side TW?

a. 3sqrt(2) (approximately 4.24)
b. 6sqrt(2) (approximately 8.49)
c. 6sqrt(3) (approximately 10.39)
d. 8
e. 9

Question

In triangle TVW below, what is the length of side TW?

a. 3sqrt(2) (approximately 4.24)
b. 6sqrt(2) (approximately 8.49)
c. 6sqrt(3) (approximately 10.39)
d. 8
e. 9

aripotta · Answer

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anonymous · Answer

@xo.minnie.xox  mum cud u help her?

opcode · Answer

This requires trigonometry.
Ever heard of SOHCAHTOA? @AriPotta

aripotta · Answer

yes that rings a bell. i heard of that like 3 yrs ago lol. haven't had to use it much so it must have slipped my mind

aripotta · Answer

not quite sure what it means though, could you explain it?

opcode · Answer

SOH: $\sin \theta = \dfrac{opposite}{hypotenuse}$
CAH: $\cos \theta = \dfrac{adjacent}{hypotenuse}$
TOA: $\tan \theta = \dfrac{opposite}{adjacent}$
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opcode · Answer

We need to find the length of the adjacent side, then we can use the Pythagorean theorem to find the length of the hypotenuse.

aripotta · Answer

so this is what i have? what do i do?
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aripotta · Answer

the 45 is in degrees btw

opcode · Answer

Wait, how did you make that triangle it should be: |dw:1416507699645:dw|