C. TriangleBased Solutions
1. Basis
The two known and one unknown points are the verticies of a plane triangle (not accounting for geodetic effects), Figure C1.
Figure C1 
Plane trigonometry can be used to solve different triangle parts until enough pieces are known to allow a Forward Computation from one known point or the other.
2. Plane triangle formulae
A plane triangle, Figure C2, has six parts: three angles and three sides.

Figure C2 
Individual angles may be acute (<90°), right (=90°), or obtuse (>90°) with their sum exactly 180°.
To geometrically define a triangle requires three parts, including at least one side, be fixed. Why a side? Because fixing angles alone does not constrain the size of the triangle. The two triangles in Figure C3 are different sizes even though they have identical angles.
Figure C3 
Equations for triangle trigonometry are:
Angle condition: 

Equation C1 
Law of Sines: 

Equation C2 
Law of Cosines: 

Equation C3 
A right triangle is a special case where one angle is exactly 90°. Applying the Law of Cosines, Equation C3, to a right triangle, Figure C4:
Figure C4 
Because cos(90°) = 0 The Law of Cosines becomes the Pythagorean Theorem. 
Depending on which parts of it are known the area of a triangle can be determined using one of two equations:
Using two sides and an included angle: 
Equation C4 

Using three sides: 
Equation C5 

where:  Equation C6 
3. Trigonometric Functions
a. Sine; Law of sines
The sine of any angle falls between 1.0 and +1.0. Figure C5 shows how angles and their sines relate. This curve repeats itself every 360°; so sin(340°) = sin(20°) = sin(380°)...
Figure C5 
Taking the arcsine (a.k.a. inverse sine, sin^{1}) of a number between 1.0 and +1.0 on a calculator will always return an angle between 90° and +90°, Figure C6.
Figure C6 90° to +90° 
Table C1 shows how angles outside ±90° relate through sin() and sin^{1}():
Table C1  
a  x = sin (a)  b = sin^{1}(x)  relationship 
30°  +0.500000  +30°  b= a 
150°  +0.500000  +30°  a = 180° b 
210°  0.500000  30°  a = 180° b 
330°  0.500000  30°  a = 360° + b 
For example sin^{1}(0.5) = 30° or 150° but it can only be one or the other for a particular triangle. A calculator will always return 30° even though the correct angle may be 150°. How do you know which is correct? It depends on the triangle.
To solve a triangle using the Law of Sines you must have either two angles and a side or two sides and an angle. Moreover, one known side must be opposite a known angle.
(1) Example 1
In triangle ABC, a = 12.4', b = 8.7', and B = 36°40'. Compute the remaining angles and side.
From the Law of Sines, Equation C2:
There are two angles whose sine is +0.85112: A = 58°20' and A = (180°00'  58°20') = 121°40', Figure C7.
Figure C7 
This means there are two possible triangles, Figure C8.
Figure C8 
Consider point C as the center of an arc of radius 8.7'. This arc can intersect the remaining side at two points; A_{1} and A_{2}, creating two different triangles.
Compute the remaining components for each triangle:
For A = 58°20', Figure C9:
Figure C9 
From the angle condition:, Equation C1:
Using Law of Sines:
For A = 121°40', Figure C10:
Figure C10 
From the angle condition
Using Law of Sines
Which triangle is the correct solution? It depends on the situation. Notice that the problem statement provided numbers but not a sketch. Had a sketch been included, then the correct triangle could have been selected.
(2) Example 2
Compute angle B in the triangle of Figure C11.
Figure C11 
The missing angle can be computed two ways:
 By Law of Sines
 By angle condition
As a result there are two “correct” answers for the missing angle;
sin(108°) = sin(72°) = 0.95105652
But only 108° fits both the Law of Sines and the Angle condition.
We can also visually determine which angle is correct, providing we have a reasonably drawn sketch. In Figure C11 angle B is larger than 90° so we would select 108°.
Be careful when using the Law of Sines to solve for an unknown angle – there could be two possible answers only one of which will fit the particular triangle. 
b. Cosine; Law of cosines
The cosine of any angle falls between 1.0 and +1.0. Figure C12 shows how angles and their cosines relate. This curve repeats itself every 360°; so cos(340°) = cos(20°) = cos(380°)... The cosine curve is identical to the sine curve except its phase differs by 90°.
Figure C12 
Taking the arccos (a.k.a. inverse cosine, cos^{1}) of a number between 1.0 and +1.0 on a calculator will always return an angle between 0° and 180°, Figure C13.
Figure C13 0° to 180° 
Table C2 shows how angles outside ±180° relate through cos() and cos^{1}():
Table C2  
a  x = cos(a)  b = cos^{1}(x) 
60°  +0.500000  +60° 
120°  0.500000  +120° 
300°  +0.500000  +60° 
240°  0.500000  +120° 
Using the Law of Cosines will not cause an ambiguous solution as does the Law of Sines since any single angle in a triangle cannot exceed 180°.
To solve a triangle using the Law of Cosines you must have either three sides, or, two sides and an angle.
(1) Example
Compute the value of the angle R in the triangle of Figure C14.
Figure C14 
From the Law of Cosines:
The Law of Cosines returns only one legitimate value when solving triangles.
c. Tangent
Unlike sine and cosine, the tangent of any angle not limited to the range of 1.0 to +1.0. As a matter of fact, the tangent range is ±(infinity). You can see that the tangent function plot, Figure C15, is not sinusoidal as are the sine and cosine plots. And unlike the other two it repeats itself every 180°;
Figure C15 
The tangent curve is asymptotic at 90°, 270°, 450°, etc. Asymptotic means the curve gets close to, but never reaches, those values. Recall that tangent is sine divided by cosine. At 90°, sine = 1.0, cosine = 0.0, so tangent = 1.0/0.0 (with an identical pattern every 180°). The result of dividing by zero is infinity, hence the asymptotic plot.
Try evaluating tan(90°) on your calculator; you’ll probably get an error statement of sorts. Then try tan(89.99999°); you should get a pretty big number.
There is a Law of Tangents, but we don't generally use it to solve triangles since the Laws of Sines or Cosines are usually sufficient .
4. Solving triangles
If we start with three triangle parts (including at least one distance), we can solve the reaming three parts using combinations of the Law of Sines and Law of Cosines.
Usually (although not always) there is only one way to start solving a triangle. With each additional part, there are more ways to compute remaining pieces.
Figures C16 and C17 show different initial angle and distance combinations and the logic to solve remaining trianle parts.
Figure C16 
Given: Sides a, b, and c Compute: (1) angle A using Law of Cosines. (2) angle B using Law of Cosines or Law of Sines. (3) angle C using Law of Cosines or Law of Sines or angle condition. 
Figure C17 
Given: Sides b, c and included angle A. 
Try figuring out the logic to solve triangles in Figures C18 through C21 on your own.

Given: Angles A, C and side b. 

Given: Angles A, C, and side a. 

Given: Sides c, b, and angle C 

Given: Angles A, B, and C 
5. Intersections
a. Solution logic
Determining the position of an unknown point requires a combination of triangle solutions along with inverse and Forward Computations. Two points with known coordinates (A and B) form one side of a triangle; inversing between them gives the length and direction of that side (the base line), Figure C22.
Fig C22 
Recall that three parts of a triangle, including one side, must be known to compute the other parts. Performing an Inverse Computation between the two known points fixes one side of the triangle. The other two parts come from the measurements connecting the unknown point with the base line. Using the Law of Sines or Cosines, enough additional triangle parts are computed to connect the unknown point to a known point with a distance and direction. The unknown coordinates can then be computed using a Forward Computation.
In each of the following intersection descriptions:
 Points J and K have known coordinates and define the base line
 P is the unknown point whose position is being determined
The first step in each intersection is to inverse between points J and K to determine the base line's length and direction, thus fixing one side of the triangle and its orIentation with respect to north. The final step is a Forward Computation from one end of the base line to the unknown point. Either end of the base line may be used, computing from the other could serve as a math check.
b. Distancedistance
Distances from points J and K to point P (d_{J} and d_{K}) are known, Figure C23.
Figure C23 Distancedistance intersection 
A distancedistance intersection results in two possible locations, Figure C24.
Figure C24 Two intersection points 
Triangles JKP_{1} and JKP_{2} are mirror images: they have identical angles and distances. It's up to the surveyor to decide which point intersection point, P_{1} or P_{2} , to be solve. This mirror triangle situation is the reason it's difficult to determine the unknown position using simultaneous solution of Equations B1 and B2. The azimuth from Point J to point P is:
Equation C7 
where J is the angle at point J between the base line and the line to point P.
It's relatively simple using triangle equations to solve for point P once we know which side of the base line it is.
We'll solve for point P_{1}, Figure C25, using a Forward Computation from point J at the base line end. We already know the distance d_{JP}, the direction Az_{JP} is needed.
Figure C25 Distancedistance solution 
Solution process
Set up the Law of Cosines and rearrange it to compute angle Q
Compute azimuth, Az_{JP} , from point J to the intersection point. Because point P is to the left of the base line, Equation C7 becomes:
Forward compute from point J
Comments
What would change had we wanted the position of point P_{2} instead of point P_{1}? Angle J would be the same except it's right of the base line instead of right. Equation C7 would be:
In order for there to be an intersection point, (d_{JP} + d_{KP}) must be greater than d_{JK}. If (d_{JP} + d_{KP}) = d_{JK} then the intersection point is on the base line and the trianlge is a straight line, Figure C26.
Figure C26 
c. Directiondistance
Direction from one control point (Az_{JP}), and distance from the other (d_{KP}) are known, Figure C27.
Figure C27 Directiondistance intersection 
Let's determine the position of point P using a Forward Computation from point J. To do so, we need the distance d_{JP}. We know two sides (d_{JK }and d_{KP}) and the angle at point J which can be calculated from the two directions Az_{JK} and Az_{JP}. The simplest way to sovle d_{JP} is using the Law of Sines which requires the angle at K.
The solution process is:
Compute angle J by finding the difference between the base line direction (Az_{JK}) and direction to Point P (Az_{JP}). The smaller direction is subtracted from the larger. In this case:
Determine the angle at P using the Law of Sines
Remember that we have to be careful using the Law of Sines to compute an angle because it is ambiguous: the angle can be P or 180°P. Figure C28 shows the two different possible triangles. Distance d_{KP} can intersect the direction Az_{JP} at two locations.
Figure C28 Directiondistance two intersections 
We have to pick point P_{1} or P_{2} based on whether the angle P is acute (<90°) or obtuse (90°) which depends on the particular situation.
Once the correct angle P is selected, angle K is computed from the angle condition
Distance d_{JP} is determined from the Law of Sines
Finally, forward compute from point J to get the coordinates
Comments
If angle P is exactly 90° a directiondistance intersection will have only a single intersection point, Figure C29.
Figure C29 Directiondistance single intersection condition 
d. Directiondirection
Directions from points Q and R to point P (Az_{Q}, and Az_{R}) are known, Figure C30.
Figure C30 
Two directions intersect at only a single point so there are no multiple solutions for this method. We'll look at computing point P's coordinates as a Forward Computation from point J.
The process is:
Compute angle J as the difference between Az_{JP} and base line azimuth Az_{JK}
Compute angle K as the difference between Az_{KP} and base line backazimuth Az_{KJ }= (Az_{JK}±180°)
Compute angle P using the angle condition.
Use Law of Sines to solve distance from point J to point P
Forward compute from J
Comments
The closer Az_{JP} and Az_{KP} values are to each other, the further away their intersection will be; if the two azimuths are parallel, then no intersection is possible, Figure C31.
Figure C31 
Directiondirection is the only method which has a single intersection point. The other two methods involve at least one distance which result in two possible intersections.