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Mathematics 17 Online
OpenStudy (here_to_help15):

@Data_LG2 @e.mccormick @inowalst

OpenStudy (here_to_help15):

OpenStudy (here_to_help15):

@UnkleRhaukus

OpenStudy (unklerhaukus):

\[\boxed{\displaystyle\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}}\] In your case: the angle \(\theta=35°\), the opposite side is \(x\), and the adjacent side is 12 cm. Solve for \(x\).

OpenStudy (here_to_help15):

So how would i solve for it?

OpenStudy (unklerhaukus):

\[\tan 35° = x/12\ \text{cm}\] Multiply both sides by 12 cm, \[x =\]

OpenStudy (here_to_help15):

Wait @UnkleRhaukus do i multiply 35 by 12 or multiply just the x by 12?

OpenStudy (here_to_help15):

@UnkleRhaukus ?

OpenStudy (here_to_help15):

@iambatman

OpenStudy (anonymous):

Multiply by 12 cm...Unkle even told you. `Multiply both sides by 12 cm`

OpenStudy (anonymous):

tan35*12

OpenStudy (here_to_help15):

Ok i multiplied that already

OpenStudy (unklerhaukus):

how many cm did you get?

OpenStudy (here_to_help15):

8.4

OpenStudy (here_to_help15):

I turned it in i thought you gave up on me ;(

OpenStudy (unklerhaukus):

\[\tan 35 = x/12\ \text{cm}\\ \tan 35\times12\ \text{cm} = x/12\ \text{cm}\times12\ \text{cm}\\ \tan 35\times12\ \text{cm} = x\]

OpenStudy (unklerhaukus):

8.4 cm \(\checkmark\), Good work

OpenStudy (here_to_help15):

thank you and @UnkleRhaukus mind helping me on this one big thing? I'd be so grateful for your assistance

OpenStudy (here_to_help15):

OpenStudy (here_to_help15):

Ok can you help me with this or ?

OpenStudy (unklerhaukus):

do you have some documentation on building codes for ramps? [because i don't].

OpenStudy (here_to_help15):

http://www.adawheelchairramps.com/wheelchair-ramps/ada-guidelines.aspx @UnkleRhaukus

OpenStudy (unklerhaukus):

1:12 is the ratio of the opposite side, to the adjacent side

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