`theta` is an angle between `0^circ` and `90^circ`. Draw a right-angled triangle with `theta` as one of the angles. Any two such triangles will be similar so the ratio of the sides of the triangle will be the same.
Can define:
Often remembered by: soh-cah-toa
sin, cos and tan are abbreviations of sine, cosine and tangent.
Need to know these triangles for exact values of sin, cos and tan with the angles `30^circ`, `45^circ` and `60^circ`.
Q: A surveyor stands 25m from a building and measures the angle to the top from the horizontal to be `30^circ`. How high is the building?
A:
An angle has size `theta` radians if the arc it subtends on the circumference of a circle of radius `1` has length `theta` units.
Circumference of circle of radius `1` is `2pi`. So we have
In general
So
| Degrees | `360^circ` | `180^circ` | `90^circ` | `60^circ` | `45^circ` | `30^circ` |
| Radians | `2pi` | `pi` | `pi/2` | `pi/3` | `pi/4` | `pi/6` |
Draw a circle radius 1. The point P which lies on the circle has coordinates `(cos theta, sin theta)`.
This allows us to give cos and sin for any angle, and we define
Q: What is the exact value of `sin 300^circ`?
A: The coordinates of the point `P` are `(cos 300^circ , sin 300^circ)` so we need to find the values for these. Draw in a triangle between the `x`-axis and the point `P`. The sides of the triangle are of length `sqrt(3)/2` and `1/2`. Hence the coordinates of `P` are `(1/2, - sqrt(3)/2)`. Therefore we have found that
To simplify the process we use the method of the associated acute angle. This is the angle that the line makes with the `x`-axis in the saded part of the diagram below on the left. We then use that angle and correct the sign using the diagram on the right.
Q: What are the exact values of `sin - 150^circ` and `tan -150^circ`?
A: Associated accute angle is `180^circ - 150^circ = 30^circ`.
`sin 30^circ = 1/2` and `tan 30^circ = 1/sqrt(3).
`-150^circ` is in the 3rd quadrant so sin is negative and tan is poisitive.
Hence `sin - 150^circ = - 1/2` and `tan -150^circ=1/sqrt(3)`.
Q:Solve `2sin theta - 1=0` for `0 le theta < 2pi`.
A: Rearrange the equation to get `sin theta = 1/2`. So `sin theta` is positive, and therefore `theta` lies in the 1st and 2nd quadrants.
Associated acute angle is `sin^(-1) 1/2 = pi/6`.
In the 1st quadrant `theta = pi/6`.
In the 2nd quadrant `theta = pi - pi/6 = (5pi)/6`.
So the solutions in the range `0 le theta < 2pi` are `theta = pi/6, (5pi)/6`.
Q: Solve `1=1/2 - sin 4x` for `0^circ le x le 150^circ`.
A: Rearrnage the equation to get `sin 4x = - 1/2`. Since `sin 4x` is negative there will be solutions in the 3rd and 4th quadrants.
The associated acute angle is `sin^(-1) = 30^circ`.
In the 3rd quadrant `4x = 180^circ + 30^circ = 210^circ`.
In the 4th quadrant `4x = 360^circ - 30^circ = 330^circ`.
Since `0^circ le x le 150^circ` we are looking for values of `4x` in the range `0^circ le 4x le 600^circ`.
So `4x = 210^circ, 330^circ, 570^circ`.
Therefore `x = 52.5^circ, 82.5^circ, 142.5^circ`.
Can only use soh-cah-toa with right-angled triangles.
But for non-right-angled triangles we have:
Q: What is the size of the angle `A` in the diagram?
A: Use the cosine rule with `a=5`, `b=8` and `c=7`. So
Therefore `A=38.2^circ qquad` (1 d.p.)