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These are right triangles that will use either sin, cos, or tan, depending upon what you have to work with in regards to the reference angle. The first one has a reference angle of 51 with y being the side opposite it and 12 being the hypotenuse. The sin identity uses the side opposite over the hypotenuse as its formula:
and 12 sin(51) = y and y = 9.325
The second one has the reference angle as the unknown. You could use any of the identities here because you have all the sides of the triangle, but I will use sin again:
and
and
The next one has a referece angle of 13 with 24 being the side adjacent to it and the unknown being the side across from it. You will use the tangent identity here:
and 24 tan(13) = x so x = 5.540
The last one has a reference angle of 20 with the hypotenuse as the unknown x, and the side across from it as 10. Use the sin identity again:
and
and
with x = 29.238
Everything is in regards to the reference angle; you HAVE to be able to identify the reference angle and then how the given sides are related to it.
and 12 sin(51) = y and y = 9.325
The second one has the reference angle as the unknown. You could use any of the identities here because you have all the sides of the triangle, but I will use sin again:
and
and
The next one has a referece angle of 13 with 24 being the side adjacent to it and the unknown being the side across from it. You will use the tangent identity here:
and 24 tan(13) = x so x = 5.540
The last one has a reference angle of 20 with the hypotenuse as the unknown x, and the side across from it as 10. Use the sin identity again:
Everything is in regards to the reference angle; you HAVE to be able to identify the reference angle and then how the given sides are related to it.
The absolute value function |<em>x</em>| always returns a non-negative number. It takes any number <em>x</em> and returns <em>x</em> if it's already non-negative, or -<em>x</em> if it is negative in order to make it positive.
For the equation
-3 + 4 |2<em>x</em> - 5| = 14
rearrange terms to get
|2<em>x</em> - 5| = 17/4
Now,
• if 2<em>x</em> - 5 ≥ 0, then |2<em>x</em> - 5| = 2<em>x</em> - 5. Then
2<em>x</em> - 5 = 17/4
• and if instead 2<em>x</em> - 5 < 0, then |2<em>x</em> - 5| = -(2<em>x</em> - 5), so that
-(2<em>x</em> - 5) = 17/4, or
2<em>x</em> - 5 = -17/4
In the first case,
2<em>x</em> - 5 = 17/4
2<em>x</em> = 17/4 + 5 = 37/4
<em>x</em> = 37/8
In the second case,
2<em>x</em> - 5 = -17/4
2<em>x</em> = -17/4 + 5 = 3/4
<em>x</em> = 3/8