MHB Solving Sine Rule Question: Angle B & A in Triangle ABC

[ai93]

I understand how to use the Sine rule, but I think I may get stuck halfway through!

My questions is:
In a Triangle ABC the angle at C is 48.15 degrees, side AB is 15.3m and side AC is 17.6m.
Calculate the size of angle B and angle A, giving all possible solutions, in degrees accurate to 1 d.p.

I used the Sine rule.
$$\frac{SinB}{17.6}=\frac{Sin(48.15)}{15.3}$$
$$\therefore$$$$SinB=\frac{15.3}{Sin(48.15)x17.6} = 0.86$$

so, B=$$Sin^{-1}(0.86)$$
B= 59 degrees.

I understand up to here, but in the solution B can = 59 or 120.9 degrees. Where I am not sure where 120.9 comes from!

Therefore I can only solve for B but not for A. Any help?

 

[MarkFL]

There are two angles for which:

$$\sin(B)\approx0.86$$ and $$0^{\circ}<B<180^{\circ}$$

You found the first quadrant angle, but recall the identity:

$$\sin\left(180^{\circ}-\theta\right)=\sin(\theta)$$

Therefore, we also find that it may be possible that:

$$B\approx180^{\circ}-\arcsin(0.86)\approx120.68^{\circ}$$

Now, since:

$$120.68+48.15=168.83<180$$

We know there are two possible triangles that satisfy the given conditions. You can read more about the ambiguous case here:

Law of sines - Wikipedia, the free encyclopedia

 

[ai93]

There are two angles for which:

$$\sin(B)\approx0.86$$ and $$0^{\circ}<B<180^{\circ}$$

You found the first quadrant angle, but recall the identity:

$$\sin\left(180^{\circ}-\theta\right)=\sin(\theta)$$

Therefore, we also find that it may be possible that:

$$B\approx180^{\circ}-\arcsin(0.86)\approx120.68^{\circ}$$

Now, since:

$$120.68+48.15=168.83<180$$

We know there are two possible triangles that satisfy the given conditions. You can read more about the ambiguous case here:

Law of sines - Wikipedia, the free encyclopedia

I fairly understand..

So, since we found out the first quadrant, angle B which is 59 degrees.
We can say B $$\approx180 - sin^{-1}(0.86)\approx120.86$$

Can I ask when do we know that the angle can have two solutions?

We have angle C as 48.15, angle B can be either 59 or 120. Or can angle A be 168.
Sorry, abit confused!

 

[MarkFL]

We know there can be two triangles because:

$120.68+48.15=168.83<180$

Because the sum of the given angle and the larger possibility is less than $180^{\circ}$, we know this leaves room for the third angle.

So, when $\angle B\approx 59.32^{\circ}$ then $\angle A\approx180^{\circ}-\left(48.15+59.32\right)^{\circ}$ and when $\angle B\approx 120.68^{\circ}$ then $\angle A\approx180^{\circ}-\left(48.15+120.68\right)^{\circ}$. This is because the sum of the three interior angles of a triangle is always $180^{\circ}$.

 

[ai93]

We know there can be two triangles because:

$120.68+48.15=168.83<180$

Because the sum of the given angle and the larger possibility is less than $180^{\circ}$, we know this leaves room for the third angle.

So, when $\angle B\approx 59.32^{\circ}$ then $\angle A\approx180^{\circ}-\left(48.15+59.32\right)^{\circ}$ and when $\angle B\approx 120.68^{\circ}$ then $\angle A\approx180^{\circ}-\left(48.15+120.68\right)^{\circ}$. This is because the sum of the three interior angles of a triangle is always $180^{\circ}$.

Thanks! Makes a lot more sense now :D