Unit 57: Complementary Angle Relationships

When presented with a right triangle as such,

Then 

\(sin\;x=cos\;y\)

and

\(cos\;x=sin\;y\)

To see why this is, we can just give our triangle some measurements.

Going back to our SohCahToa, sin of x = \(\frac OH=\frac35\)   By the same token, cos y = \(\frac AH=\frac35\)

Another identity to be aware of is

$$sin\;x=cos(90-x)$$

Going back to our triangle, we have a 90 degree angle, and if x is arbitrarily 33°, we would find by subtracting 90 – 33 (because \(x+y\) must equal 90°).  So 90 – x is the same as y.  

Here’s a rather tricky example:

The angles shown above are acute and sin(a°) = cos(b°). If \(a=4k-22\) and \(b=6k-13\), what is the value of 

A) 4.5
B) 5.5
C) 12.5
D) 21.5

Strategy:

  • Always remember that if the \(\boldsymbol s\boldsymbol i\boldsymbol n\mathbf{\left({some\;angle}\right)}\boldsymbol=\boldsymbol c\boldsymbol o\boldsymbol s\boldsymbol(\boldsymbol a\boldsymbol n\boldsymbol o\boldsymbol t\boldsymbol h\boldsymbol e\boldsymbol r\boldsymbol\;\boldsymbol a\boldsymbol n\boldsymbol g\boldsymbol l\boldsymbol e\boldsymbol)\), then the two angles inside of the parentheses always add up to be 90 (no matter what they look like).

Back to our example:

It might not be obvious at first, but by telling us sin (a°) = cos (b°), we know that these two angles must add up to 90.

$$a+b=90$$

Now substitute in for a and b:

$$a=4k-22\;\;\;\;\;\;\;\;\;\;\;b=6k-13\\\left(4k-22\right)+\left(6k-13\right)=90\\10k-35=90\\10k=125\\k=12.5$$

Again, this one is tricky.  If you need to, walk back through the steps again until it’s clear.