Further Maths - literally spent hours I think I need fresh eyes to see my flaw >.<

[DeadlyKamina]

Further Maths - literally spent hours I think I need fresh eyes to see my flaw >.<

A triangle has a length of squareroot 3, its opposite angle is theta. the next length is the squareroot of 6, opposite angle of 45deg.

So far I've tried the sine rule but to no avail working out root 6 over sine45 as root 12 (or 2 * root 3)

Please help

Much appreciated.

*ADDED*
I should have said but its a non-calculator question so no use of inverse sine >..< So far as the trianlgle is scalene I decided to draw a perpendicular line from the top angle, this split the triangle into a scalene and a isosceles, the isosceles had a hypotenuse of root 3 and so both other sides were length cuberoot 3, this left the scalene with a hypotenuse of root 6, and lengths cuberoot 3 and root(6-root3) although I am unsure if these lengths are important All I know for sure is the sine rule should play some part of it, root6/sine(45) = root3/sine(x). I will also add we are given the values of sin, tan and cos of 45, 60 and 30

 

[stapel]

A triangle has a length of squareroot 3, its opposite angle is theta. the next length is the squareroot of 6, opposite angle of 45deg.

[I've] literally spent hours I think I need fresh eyes to see my flaw: So far I've tried the sine rule but to no avail working out root 6 over sine45 as root 12 (or 2 * root 3)

Please post your work, so that we can try to find any flaws. Thank you!

 

[DeadlyKamina]

I should've said no calculators too so no inverse of sine >...< my working so far is root 6 over sine45 is equal to root 3 over sine(x) i need to solve for x, I tried drawing a perpendicular line from one corner so I had two triangles, one isosceles, hypotenuse length of root 3 and the other two sides at cuberoot 3, the second triangle, scalene, had lengths of cuberoot three, root 6 and root 6-root3, although I'm unsure if these lengths are required for a solution

 

[DeadlyKamina]

I should have said but its a non-calculator question so no use of inverse sine >..< So far as the trianlgle is scalene I decided to draw a perpendicular line from the top angle, this split the triangle into a scalene and a isosceles, the isosceles had a hypotenuse of root 3 and so both other sides were length cuberoot 3, this left the scalene with a hypotenuse of root 6, and lengths cuberoot 3 and root(6-root3) although I am unsure if these lengths are important All I know for sure is the sine rule should play some part of it, root6/sine(45) = root3/sine(x). I will also add we are given the values of sin, tan and cos of 45, 60 and 30

 

[DeadlyKamina]

Please post your work, so that we can try to find any flaws. Thank you!

see update <3

 

[stapel]

I should've said no calculators too so no inverse of sine >...< my working so far is root 6 over sine45 is equal to root 3 over sine(x) i need to solve for x...

Since "forty-five degrees" is one of the basic reference angles for which you've memorized the values (here), you don't need a calculator. Just plug in the memorized value, and solve the proportion.

 

[DeadlyKamina]

Since "forty-five degrees" is one of the basic reference angles for which you've memorized the values (here), you don't need a calculator. Just plug in the memorized value, and solve the proportion.

so root 12 = root 3/sine(x)

so I need sine(x) to be 1/root4 no? so root3/1/root4 to get root 12?

 

[Ishuda]

so root 12 = root 3/sine(x)

so I need sine(x) to be 1/root4 no? so root3/1/root4 to get root 12?

What is root4?


From original post

A triangle has a length of squareroot 3, its opposite angle is theta. the next length is the squareroot of 6, opposite angle of 45deg. ...
I will also add we are given the values of sin, tan and cos of 45, 60 and 30

What stapel said but in a little more detail: Use the Law of Sines
\(\displaystyle \dfrac{sin(\theta)}{\sqrt{3}}\, =\, \dfrac{sin(45^\circ)}{\sqrt{6}}\, =\, \dfrac{1}{2\, \sqrt{3}}\)
or
\(\displaystyle sin(\theta)\, =\, \dfrac{1}{2}\)
From your table, look up \(\displaystyle \theta\). Now that you know two angles and remembered that a triangle has a total of 180\(\displaystyle ^{\circ}\), you can compute the third angle. Use the Law of Sines and your table again to compute the third side.

 

[DeadlyKamina]

What is root4?


From original post

What stapel said but in a little more detail: Use the Law of Sines
\(\displaystyle \dfrac{sin(\theta)}{\sqrt{3}}\, =\, \dfrac{sin(45^\circ)}{\sqrt{6}}\, =\, \dfrac{1}{2\, \sqrt{3}}\)
or
\(\displaystyle sin(\theta)\, =\, \dfrac{1}{2}\)
From your table, look up \(\displaystyle \theta\). Now that you know two angles and remembered that a triangle has a total of 180\(\displaystyle ^{\circ}\), you can compute the third angle. Use the Law of Sines and your table again to compute the third side.

Thank you so much that was really dumb of me really sorry for wasting your time I cant believe I didn't realise >...< I swear to you all I do the further maths course for a reason but sometimes I do miss the obvious >..< really thank you very much that was so stupid of me xD