The ambiguous case of the sine law, in solving triangle problems, is explored interactively using an applet.

Review
The sine law is given by

sin(a)/a = sin(b)/b = sin(g)/c

Let us solve the triangle problem when we are given sides a and b and angle a, with a acute.

The algebraic solution to this problem starts by finding sin(b) which is given by

sin(b)=b*sin(a)/a

Then find angles b and g and side c.

For the triangle to exist sin(b) must be less than or equal to 1.

There is also a GEOMETRIC solution to this problem which we describe below.

step 1: draw a line (l) and draw point A on this line.

step 2: draw a segment AC that makes an angle a with line (l). Segment AC has length b.

step 3: Draw a circle with center at C and radius equal to side a and locate the points of intersection of the circle and line (l). Three cases are possible.

case 1: if a is smaller than h ,the circle does not intersect line (l) and the problem has no solution.

case 2: if a is equal to h, the circle touches line (l) at 1 point and the problem has one solution. Triangle ABC is a right triangle.

case 3: if a is larger than h, but smaller than b, the circle intersects line (l) at 2 points and the problem has 2 solutions: Triangle ABC and Triangle AB'C. This is called the ambiguous case.

case 4: if a is larger than or equal to b, the problem has one solution.

step 4: draw the third side of the triangle(s) by joining C and B, and C and B'. The figures below show the case with two solutions.

Interactive Tutorial

click on the button above "click here to start" and MAXIMIZE the window obtained.

Use the sliders to set parameters b and alpha to 3 and 30 degrees respectively if they are not already. Use the red point on the circle to increase or decrease the radius a of the circle.

Find a value of a that correspond to a case with no solution as described above and check by calculations.

Find a value of a that correspond to a case with 1 solution as described above and check by calculations.(this may be a little difficult to do , but a close value should be enough).

Find a value of a that correspond to a case with 2 solutions as described above and check by calculations.

Find a value of a that correspond to a case with 1 solution as described above and check by calculations.

Set b and alpha to other values and explore further.