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Mathematics 7 Online
OpenStudy (hang254):

In △DEF, what is the length of DF ? 90 45√2 15√3 30√3

OpenStudy (hang254):

OpenStudy (callisto):

Use sine ratio to find DF \[sin60 = \frac{45}{DF}\]

OpenStudy (hang254):

(45)sin60 = 38.97114317

OpenStudy (callisto):

Not really.

OpenStudy (hang254):

so then???

OpenStudy (callisto):

Multiply both sides by DF. Then divide both sides by sin60. What do you get?

OpenStudy (hang254):

I dont know what DF equals

OpenStudy (ganpat):

ok... Sin 60 = 45/ DF Therefore DF = 45/ sin 60 = 45/\[\sqrt{3}/2\] = 90/ sq rt 3...

OpenStudy (hang254):

oh, ok. So 51.96

OpenStudy (hang254):

so, 30 sqrt 3

OpenStudy (ganpat):

yup.. 51.96.. if u simplify further...

OpenStudy (callisto):

Hmm You'll now what DF equal to \[sin60=\frac{45}{DF}\]Multiply both sides by DF. \[DFsin60=DF(\frac{45}{DF})\]\[DFsin60=45\] Then divide both sides by sin60. \[\frac{DFsin60}{sin60}=\frac{45}{sin60}\]\[DF=\frac{45}{sin60}=...\]

OpenStudy (hang254):

Thank you both :)

OpenStudy (callisto):

\[\frac{45}{sin60} = \frac{45}{\frac{\sqrt3}{2}} = \frac{90}{\sqrt3} = \frac{90\sqrt3}{3} = 30\sqrt3\]

OpenStudy (ganpat):

perfect.. multiply and divide by root 3.. denominator comes to be 3.. divide 90 by it.. So 30 rt 3...

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