Can you work this out (help much appreciated)...?

Question

Geometry and Trigonometry grade 11

Use the figure to prove that
sin 15°=sqrt6-sqrt2/4

Note: I have also been given a right-angled triangle. It has been split into 2 triangles diagonally to the left (ie. One of them kept the initial right-angle and one isn't a right angled triangle. It was split (the large right-angled triangle) into 2 small triangles at the horizontal. It (large triangles is labeled A (far left, horizontal) next to a 30° angle, C at the point of the large triangle (or where the two triangles meet highest part of the triangle in height), B which is next to the right-angle on the horizontal and D which is where the large triangle got split up on the horizontal near 1/2 of the length of the horizontal (not quite). Between C and B (vertical line) is the same length as B and D (on the horizontal, far right).

sqrt=square root. I am not given any actual lengths of the large triangle. I am given 30° next to A and a right-angle.

Thankyou so much for helping.

Answer 1

What do you mean by "diagonally"? There is no diagonal in a triangle.

Answer 2

I think I understand your diagram: you have a right-angled triangle ABC with A = 30° and B = 90°, and a point D between A and B such that CD = BC.

Note that angles BDC and BCD are both 45°, since BCD is an isosceles triangle and angle B is 90°. Also, angle ACB is 60° and so angle ACD must be 15°.

Let BC = x. Then BD = x also, and since sin 30° = 1/2, AC = 2x. From the Pythagoras theorem we get CD = x√2, and AB = √((2x)^2 - x^2) = x√3. So AD = AB - BD = x(√3 - 1).

Now, applying the sine rule to triangle ACD gives us sin 15° / (x(√3 - 1)) = sin 30° / (x√2)
and hence
sin 15° = (√3 - 1) . (1/2) / √2
= √2 (√3 - 1) / 4
= (√6 - √2) / 4.