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Wize High School Grade 11 Math Textbook > Trigonometry with Non-Right Triangles

Sine Law - Ambiguous Case

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Sine Law - The Ambiguous Case

Try it out!

Draw as many triangles as you can with one side that is 10 cm and another side that is 6 cm, where the angle across from the 6 cm side is 33°33\degree.



Since we are able to draw more than one triangle with these given measurements, we say that this is ambiguous.

When we use the sine law to find the missing side and angle measurements in this triangle, we will end up with 2 possibilities, this is called the ambiguous case of the sine law.
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Summary

When we are given 2 sides of a triangle (aa and bb) and one of the opposite angles (∠A\angle A is acute), we have the following cases:

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Example: Sine Law - The Ambiguous Case

Tom lives in an apartment building where a cat has been lost, and Tom knows it is on someone else's balcony.
A firefighter arrives with a 10m ladder and sets it up to reach Tom where it makes a 20°20\degree angle with the wall.
Tom yells down: "The cat isn't here! You are exactly 15m away from it."
How far above or below Tom should the firefighter look?

Above Triangle

To start, we can find angle bb since we can see that
20°+b=180°    ⟹    b=160°20\degree+b=180\degree \ \ \implies\ \ b=160\degree

We can use Sine law to find angle aa:
sin⁡a10=sin⁡160°15    ⟹    a=sin⁡−1(10sin⁡160°15)≈13.18°\dfrac{\sin a}{10}=\dfrac{\sin 160\degree}{15} \ \ \implies\ \ a = \sin^{-1}\left( \dfrac{10\sin160\degree}{15} \right) \approx 13.18\degree

Now that we know 2 of the 3 angles:
13.18°+160°+c=180°    ⟹    c=6.82°13.18\degree +160\degree+c= 180\degree \ \ \implies\ \ c=6.82\degree

Lastly, we can find the vertical distance using the Sine law again:
Csin⁡6.82°=15sin⁡160°    ⟹    C=15sin⁡6.82°sin⁡160°≈5.21 m  above Tom\dfrac{C}{\sin 6.82\degree}=\dfrac{15}{\sin 160\degree} \ \ \implies\ \ C = \dfrac{15\sin 6.82\degree}{\sin 160\degree} \approx \boxed{5.21 \ \rm m} \ \text{ above Tom}

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Below Triangle

We can use Sine law to find angle ee:
sin⁡e10=sin⁡20°15    ⟹    e=sin⁡−1(10sin⁡20°15)≈13.18°\dfrac{\sin e}{10}=\dfrac{\sin 20\degree}{15} \ \ \implies\ \ e = \sin^{-1}\left( \dfrac{10\sin20\degree}{15} \right) \approx 13.18\degree

Notice that this is the same as angle aa found above!
Now that we know 2 of the 3 angles:
13.18°+20°+f=180°    ⟹    f=146.82°13.18\degree +20\degree+f= 180\degree \ \ \implies\ \ f=146.82\degree

Lastly, we can find the vertical distance using the Sine law again:
Fsin⁡146.82°=10sin⁡13.18°    ⟹    F=10sin⁡146.82°sin⁡13.18°≈24 m  below Tom\dfrac{F}{\sin 146.82\degree}=\dfrac{10}{\sin 13.18\degree} \ \ \implies\ \ F = \dfrac{10\sin 146.82\degree}{\sin 13.18\degree} \approx \boxed{24 \ \rm m} \ \text{ below Tom}



Practice: Sine Law - The Ambiguous Case

Alice and Bob are looking up at the CN Tower in Toronto. Alice is 600m from the tip of the tower, and Bob is 700m from the tip, and together they form a straight line through the CN Tower.

If the angle of elevation from Bob's perspective is 52°52\degree, how far apart are Alice and Bob?