11.4.1. The Sine Rule

Given that , cm, and cm, find all the possible solutions of a triangle approximating measure of angles to the nearest second and length of sides to the nearset hundredth, if it exists.

• Acm, , or cm, ,
• Bcm, , or cm, ,
• Ccm, ,
• Dcm, ,

Example

Given that , cm, and cm, find all the possible solutions of a triangle approximating measure of angles to the nearest second and length of sides to the nearset hundredth, if it exists.

Solution

Remember the sine rule: for a triangle ,

We have been given two sides and the measure of a non-included angle. Therefore, there are three possible cases:

1. There are no triangles that satisfy the conditions.
2. There is one triangle that satisfies the conditions.
3. There are two triangles that satisfy the conditions.

Since we have been given an acute angle, we can compare the length of with the height of the triangle given by .

1. If , there are no triangles that satisfy the conditions.
2. If , there is one triangle that satisfies the conditions.
3. If , there are two triangles that satisfy the conditions.

For the conditions that have been given, to the nearest hundredth. Consequently, , therefore, we have two possible triangles which satisfy the conditions represented by the acute and obtuse solutions to the sine of the angle.

To find the possible measures of the angle, we apply the sine rule.

The first solution is when , and the second solution is when .

First Solution: when . Firstly, we calculate the measure of angle .

Now we calculate the length of using the sine rule.

Second Solution: when . Firstly, we calculate the measure of angle .

Now we calculate the length of using the sine rule.

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