Question

determine the number of triangles abc possible with the given parts. a = 42° a = 8.1 b = 10.5
how many possible solutions does this triangle have?

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that $\frac{\sin A}{a} = \frac{\sin B}{b}$. We can solve for $\sin B$:
$\sin B = \frac{b \sin A}{a}$
Substitute the given values: $A = 42^\circ$, $a = 8.1$, $b = 10.5$
$\sin B = \frac{10.5 \sin 42^\circ}{8.1}$
First, calculate $\sin 42^\circ \approx 0.6691$
Then, $\sin B = \frac{10.5 \times 0.6691}{8.1} \approx \frac{7.02555}{8.1} \approx 0.8673$

Step2: Analyze the value of $\sin B$

We know that the range of the sine function is $[-1, 1]$, and $0.8673$ is within this range. Now, we check the possible angles for $B$.
Since $\sin B \approx 0.8673$, $B$ could be in the first quadrant or the second quadrant.
First, find the reference angle: $B_1 = \arcsin(0.8673) \approx 59.9^\circ \approx 60^\circ$
Then, the other possible angle is $B_2 = 180^\circ - 60^\circ = 120^\circ$

Now, we need to check if these angles are valid with the given angle $A = 42^\circ$.
For $B_1 \approx 60^\circ$: The sum of angles in a triangle is $180^\circ$, so $C = 180^\circ - 42^\circ - 60^\circ = 78^\circ$, which is valid (positive angle).
For $B_2 = 120^\circ$: $C = 180^\circ - 42^\circ - 120^\circ = 18^\circ$, which is also valid (positive angle).

But we also need to check the side lengths. We know that $a = 8.1$ and $b = 10.5$, so $b > a$. In a triangle, the larger side is opposite the larger angle. Since $b > a$, $B$ should be larger than $A$. Both $60^\circ$ and $120^\circ$ are larger than $42^\circ$, so both are possible. Wait, but wait, let's re - check the calculation of $\sin B$.

Wait, let's recalculate $\sin B$ more accurately. $\sin 42^\circ\approx0.6691306064$
$10.5\times\sin42^\circ = 10.5\times0.6691306064 = 7.025871367$
$\sin B=\frac{7.025871367}{8.1}\approx0.867391527$

Now, let's check the angle sum for $B = 120^\circ$: $A + B=42^\circ+ 120^\circ = 162^\circ<180^\circ$, so $C = 18^\circ$, which is valid.
For $B = 60^\circ$: $A + B=42^\circ+60^\circ = 102^\circ<180^\circ$, so $C = 78^\circ$, which is valid.

But wait, another way to check is using the "ambiguous case" (SSA) rules.
The formula for the height $h$ of the triangle with respect to angle $A$ is $h = b\sin A$
$h = 10.5\times\sin42^\circ\approx10.5\times0.6691\approx7.02555$
We know that $a = 8.1$ and $h\approx7.02555$ and $b = 10.5$
Since $h < a < b$, in the SSA ambiguous case, when $h < a < b$, there are two possible triangles.

Wait, let's recall the SSA (side - side - angle) cases:
- If $a > b$: Then $A > B$, and since $A$ is acute, there is one triangle.
- If $a = h$ ($h = b\sin A$): There is one right - angled triangle.
- If $h < a < b$: There are two triangles.
- If $a < h$: There are no triangles.

Here, $h = b\sin A=10.5\times\sin42^\circ\approx7.026$, $a = 8.1$, $b = 10.5$
Since $h\approx7.026 < a = 8.1 < b = 10.5$, so there are two possible triangles.

Wait, but earlier when we calculated $\sin B\approx0.8673$, the two possible angles for $B$ are approximately $60^\circ$ and $120^\circ$, and both lead to valid triangles. So the number of possible solutions is 2.

Wait, but let's check again. Let's use the Law of Sines formula correctly.

$\frac{\sin A}{a}=\frac{\sin B}{b}\Rightarrow\sin B=\frac{b\sin A}{a}$

We have $A = 42^\circ$, $a = 8.1$, $b = 10.5$

$\sin B=\frac{10.5\times\sin42^{\circ}}{8.1}$

$\sin42^{\circ}\approx0.6691$

$\sin B=\frac{10.5\times0.6691}{8.1}=\frac{7.02555}{8.1}\approx0.8673$

Now, $\arcsin(0.8673)\approx60^{\circ}$ (approximate value). The other angle is $180 - 60=120^{\circ}$

Now, check if $A + B<180^{\circ}$ for both cases:

Case 1: $B = 60^{\circ}$, $A + B=42 + 60 = 102^{\circ}<180^{\circ}$, so $C = 78^{\circ}$ (valid)

Case 2: $B = 120^{\circ}$, $A + B=42+120 = 162^{\circ}<180^{\circ}$, so $C = 18^{\circ}$ (valid)

Also, since $b > a$, $B > A$, and both $60^{\circ}$ and $120^{\circ}$ are greater than $42^{\circ}$, so both triangles are possible.

So the number of possible solutions is 2.

Answer:

2