I must admit to having reservations as I write this blog post. Not because I am unsure as to the approaches I will outline, but rather to do with the categorisation as a 'Method of Last Resort'. Before now I have typically suggested that methods of last resort should be the things we do when our understanding of a situation doesn't allow us to take a more efficient approach - for example considering order of operations a 'Method of Last Resort' as it is the sort of thing we consider when we can't simply work left to right (as in 5 x 6 ÷ 10 for example) or when we can't simplify a calculation (as in 23 x 6 + 7 x 6 = 30 x 6, or 17

I am going to propose that SOHCAHTOA is a method of last resort. By SOHCAHTOA I don't mean the mnemonic, I mean the idea of treating the trigonometric ratios as formulae:

^{2}– 3^{2}). I am not completely sure that what I am going to outline falls into that category, but nonetheless here goes...I am going to propose that SOHCAHTOA is a method of last resort. By SOHCAHTOA I don't mean the mnemonic, I mean the idea of treating the trigonometric ratios as formulae:

So what is the alternative? Well the obvious one is the unit circle, but that might be a bit much for the first introduction of trigonometry. Instead I wanted to outline an approach around similar triangles.

Let us first take sine. Sine of an angle between 0 and 90 relates the opposite to the hypotenuse in the following way:

This is all the basis we need to find missing sides in any right triangle with the angle θ. Consider now the triangle below:

This triangle is an enlargement of the first triangle, using a scale factor of 13. This implies that the opposite side is simple 13 × sin θ. Even looking at the triangle below:

This triangle is still an enlargement of the first triangle, but with a scale factor of 13/sin θ. So the hypotenuse must by 13/sin θ.

This approach also be used to find angles. Consider the triangle below:

This triangle is still an enlargement of the original triangle, again by scale factor 13. This would mean that the opposite side of the smaller triangle is 5/13. But remember, in the smaller triangle the opposite side is sin θ. So we have that sin θ = 5/13. This leads of course to θ = sin

^{-1}(5/13).
Virtually identical approaches can be used with reference to the adjacent and hypotenuse sides, with the cosine function and the tangent function with the opposite and adjacent sides. Importantly, this approach arguably requires a deeper understanding of how trigonometric functions relate sides of a triangle than the formulae provided at the beginning of this blog post, and it is for this reason why I wonder if the formulae couldn't be considered a 'Method of Last Resort.'