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

  • Thread starter Thread starter ai93
  • Start date Start date
  • Tags Tags
    Sine
AI Thread Summary
In triangle ABC, with angle C at 48.15 degrees, side AB measuring 15.3m, and side AC measuring 17.6m, the Sine Rule is applied to find angle B, which can be either 59 degrees or approximately 120.68 degrees. The ambiguity arises because the sine function has two possible angles for a given sine value in the range of 0 to 180 degrees. When angle B is 59 degrees, angle A can be calculated as 180 degrees minus the sum of angles B and C. Conversely, if angle B is 120.68 degrees, angle A is determined similarly, confirming the existence of two possible triangles for the given conditions. Understanding the ambiguous case of the Sine Rule is crucial in solving such problems.
ai93
Messages
54
Reaction score
0
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?
 
Mathematics news on Phys.org
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
 
MarkFL said:
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!
 
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 $180^{\circ}$.
 
MarkFL said:
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top