Consider two right triangles with a common acute angle and use the Angle-Angle (AA) Similarity Theorem. Then, rewrite the proportions in terms of the sine and cosine of the common acute angle.
See solution.
Practice makes perfect
Let's consider a right triangle ABC, and let's mark one of the acute angles.
By taking ∠A as a reference point, we have that BC is the opposite side and AC is the adjacent side.
Next, we can find the sine and cosine of ∠A by finding the following ratios.
sin A = opp/hyp = BC/AB [0.2cm]
cos A = adj/hyp = AC/AB
Next, let's consider a second right triangle APQ such that ∠A is common for both triangles.
As we can see, ∠ACB ≅ ∠AQP because they are both right angles. Then, by the Angle-Angle (AA) Similarity Theorem, △ ABC ~ △ APQ.
â–³ ABC ~ â–³ APQ
⇕
AC/AQ = AB/AP = BC/PQ
Next, let's separate the equation above as follows.
AC/AQ = AB/AP = BC/PQ
↙ ↘
AC/AQ = AB/AP AB/AP = BC/PQ
Next, we will rewrite the two equations above by using the sine and cosine ratios.
ccc
AC/AQ = AB/AP & & AB/AP = BC/PQ
⇕ & & ⇕
AC/AB_(cos A) = AQ/AP & & PQ/AP = BC/AB_(sin A)
That way, we've defined the sine and cosine of ∠A by using the side lengths of △ AQP. This implies there is not an unique right triangle that must be used.
