﻿ When there are No Triangles , Ambiguous Case of Law of Sines with Zero Solutions, explained with example problems.
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# Ambiguous case of the Law of sines

## Explained with examples

#### Example of Zero Triangles Possible

Use the Law of Sines to solve for $$m \angle A$$
Triangle 1
Given
$$BC = 23 \\ AC = 3 \\ \angle B = 44^{\circ}$$.
Work

$\frac{sin( A)}{ 23 } = \frac{sin(44)}{3} \\ sin( A) = \frac{23 \cdot sin(44)}{3} \\ sin( A) = \red { 5.32571417} \\ \boxed{\text{ No Solution}}$

As you can see $$\angle A$$ is 'impossible' because the sine of an angle cannot be equal to 5.3. (Remember the greatest value that the sine of an angle can have is 1)

#### Visual of Zero Triangles Possible

Look what happens when you try to draw a line from B to A at a $$44^{\circ}$$ angle.

# There is no way to do it

### Example 1

For $$\triangle ABC$$, $$a = 6, b =10$$, and $$m \angle A = 42^{\circ}$$. How many Triangles can be formed?

Work

$$\frac{sin ( B )}{b} = \frac{sin ( A )}{a} \\ \frac{sin (\red B)}{10} = \frac{sin (42^{\circ})}{6} \\ sin(\red B)= \frac{10 \cdot sin (42^{\circ})}{6} \\ sin(\red B)= 1.11522 \\ \boxed{\text{No Solution}}$$

Zero Triangles . The maximum value of the sine function is 1.

$$sin(B)= \cancel {\red {1.11522}}$$

##### Practice Problem

Work

$$\frac{sin ( E )}{e} = \frac{sin ( F )}{f} \\ \frac{sin (\red E)}{27} = \frac{sin (37^{\circ})}{12} \\ sin(\red E)= \frac{27 \cdot sin (37^{\circ})}{12} \\ sin(\red B)= 1.3541 \\ \boxed{\text{No Solution}}$$

There is no possible $$\angle E$$ with the given dimensions . The maximum value of the sine function is 1.
$$sin(B)= \cancel {\red {1.3541}}$$