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Mathematics 18 Online
OpenStudy (anonymous):

A right triangles hypotenuse is 20m long. What is the length of the side opposite a 60degree angle? Give your answer to the nearest tenth of a cenimeter.

OpenStudy (anonymous):

@Momo_ please help? im really lost :(

OpenStudy (anonymous):

@Kwamgreen101 please help?

OpenStudy (anonymous):

@kc_kennylau

OpenStudy (kc_kennylau):

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OpenStudy (kc_kennylau):

Do you remember the definitions of sine, cosine and tangent?

OpenStudy (anonymous):

no this whole question confuses me :/ just started at a new school and im lost :/

OpenStudy (kc_kennylau):

|dw:1396756116127:dw|

OpenStudy (kc_kennylau):

Remember these definitions; they'd be very useful

OpenStudy (kc_kennylau):

So can you apply the definition of sine to the triangle in question?

OpenStudy (anonymous):

?/20 ??

OpenStudy (kc_kennylau):

Yes, but don't miss out the left hand side

OpenStudy (kc_kennylau):

What is ?/20 the value of?

OpenStudy (anonymous):

b/c or a leg/hypotenuse. right?

OpenStudy (kc_kennylau):

Okay maybe I'm confusing, I meant sinθ=?/20

OpenStudy (anonymous):

oh sorry :/

OpenStudy (kc_kennylau):

Nah it's my fault, continue?

OpenStudy (anonymous):

yes please :)

OpenStudy (kc_kennylau):

Remember the value of θ?

OpenStudy (anonymous):

no im sorry :(

OpenStudy (kc_kennylau):

Don't apologize it's not your fault; look at the first picture I gave you?

OpenStudy (anonymous):

okay 60?

OpenStudy (kc_kennylau):

So you now have \(\sin60^\circ=\dfrac?{20}\)

OpenStudy (kc_kennylau):

Do you have the special values table?

OpenStudy (kc_kennylau):

This is the table if you don't have

OpenStudy (anonymous):

thank you. so what do I do now

OpenStudy (kc_kennylau):

Solve the equation :)

OpenStudy (anonymous):

it says 60degrees = pie/3. how do I do this??

OpenStudy (kc_kennylau):

Look at the table to find the value of \(\sin60^\circ\) first?

OpenStudy (kc_kennylau):

Ignore the row with \(\pi\)s

OpenStudy (anonymous):

squareroot3/2

OpenStudy (kc_kennylau):

Yep, plug it in \(\sin60^\circ=\dfrac?{20}\)

OpenStudy (anonymous):

60deg=sqrt3/2/20

OpenStudy (anonymous):

right?

OpenStudy (kc_kennylau):

Please treat \(\sin60^\circ\) as one identity

OpenStudy (kc_kennylau):

You just said \(\sin60^\circ=\dfrac{\sqrt3}2\)

OpenStudy (anonymous):

okay so that's the equation? no 20?

OpenStudy (kc_kennylau):

No

OpenStudy (kc_kennylau):

\[\begin{array}{rcl} \sin60^\circ&=&\frac?{20}\\ \frac{\sqrt3}2&=&\frac?{20}\\ \end{array}\]

OpenStudy (kc_kennylau):

Continue

OpenStudy (anonymous):

I give up haha I'm just going to pick an answer cause this is making no sense whatsoever

OpenStudy (anonymous):

thank you for helping though

OpenStudy (kc_kennylau):

okay...

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