So I got a question. I'm working on right triangle trigonometry and I'm a little stuck on how to find the value of x when there is a side of 12 and an angle of 26deg. How would I go about solving this?
memorize these trig ratios in terms of a right-angled triangle
with sides called adjacent, opposite and hypotenuse in terms of the
base angle A:
SinA = opposite/hypotenuse or y/r
CosA - adjacent/hyptenuse or x/r
tanA = opposite/adjacent or y/x
the 3 secondary ratios are
CscA = 1/SinA = r/y
SecA = 1/CosA = r/x
cotA = 1/tanA = x/y
Many students learn it as SOHCAHTOA. Notice how the letters match
the definitions.
so you are involving x, and "a side of 12"
so the 12 should be the opposite, or else we would have called it the
hypotenuse.
so in our list above, the ratio involves adjacent and opposite, so
we must be talking about tan
tan 26° = opposite / adjacent = 12/x
x tan26 = 12
x = 12/tan26
pull out your calculator, make sure it is set on DEG using the DRG button
and enter:
12
÷
tan 26
=
you should get appr 24.6 <--- base of your right-angled triangle