Question

Could someone please help me with this question? I know I have to use the sine rule, but now I'm stuck...

In a triangle with side a= (x+3) and side b = (2x-5) with angle A = 30 degrees and angle B = 45 degrees find the exact value of x. Express your answer in the form a+b*sqrt2 where a,b =Q

Answer 1

\[\frac{ a }{ sinA }=\frac{ b }{ sinB }\]

Answer 2

substitute your stuff in

then sinB will have a root 2 in it, multiply by sinB

multiply by sinA
then solve for x

Answer 3

\[\frac{ x+3 }{ \sin30 }=\frac{ 2x-5 }{ \sin45 }\]

Answer 4

\[(x+3)\sin45=(2x-5)\sin30\]

Answer 5

do I change sin30 and sin45 to the exact values now or do I leave it and just multiply?

Answer 6

yes just use the fraction form of sin30 and sin 45

Answer 7

I got \[x=5+3*\sqrt{2}\div \sqrt{2}\]

Answer 8

how do I get rid of the sqrt2 in the denominator?

Answer 9

oh wait is the answer 6+5*sqrt2 ?

Answer 10

\[\sin45=\frac{ \sqrt{2} }{ 2 }\]
\[\sin30=\frac{ 1 }{ 2}\]
there fore we have

\[\frac{ \sqrt{2} }{ 2 }(x+3)=\frac{ 1 }{ 2 }(2x-5)\]

multiply by 2 to both sides
\[\sqrt{2}(x+3)=2x-5\]

multiply out and put x terms to the left and the constant to the right

\[\sqrt{2}x-2x=-5-3\sqrt{2}\]
factorise

\[(\sqrt{2}-2)x=-5-3\sqrt{2}\]

\[x=\frac{ -5-3\sqrt{2} }{ \sqrt{2}-2 }\]