triangle side problem

[eric beans]

what i did was pr/pq = 1/2,
also sin of angle p = 6^(1/2)/pq = 3^(1/2)/2 ,
solving for pq = 2,

so replacing 2 = pq in pr/pq = 1/2 , gives pr =1.

but the answer says it's 2. what am i doing wrong?

 

[nasi112]

[MATH]\frac{PR}{PQ} = \frac{2}{1+\sqrt{3}}[/MATH], not [MATH]\frac{1}{2}[/MATH]
if you want to find [MATH]PR[/MATH] or [MATH]PQ[/MATH], use the law of sines

[MATH]\frac{PR}{\sin 45^{o}} = \frac{\sqrt{6}}{\sin 60^{o}}[/MATH]
Solve for [MATH]PR[/MATH]
And for [MATH]PQ[/MATH]
[MATH]\frac{PQ}{\sin 75^{o}} = \frac{\sqrt{6}}{\sin{60^{o}}}[/MATH]
Solve for [MATH]PQ[/MATH]

 

[eric beans]

where'd you get this though?

PRsin45o=√6sin60oPRsin⁡45o=6sin⁡60o\displaystyle \frac{PR}{\sin 45^{o}} = \frac{\sqrt{6}}{\sin 60^{o}}

how'd you get this?

PRPQ=21+√3PRPQ=21+3\displaystyle \frac{PR}{PQ} = \frac{2}{1+\sqrt{3}}

 

[lev888]

where'd you get this though?



how'd you get this?

"use the law of sines"

 

[lex]

Yes, the simplest solution is to use the sine rule:

They give you that cos(60)=1/2, so I wonder if you have learned the cosine rule? It could be done that way too.

Or just use ordinary trigonometry