Solution of Triangles (Ambiguous Cases) : IIT JEE Maths

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

  • If the three sides a, b,  and c are given, angle A is obtained from \tan{(A/2)}= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} or \cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2bc}. B and C can be obtained in a similar way.
  • If two sides b and c and the included angle A are given, then \tan{\frac{B-C}{2}}=\frac{b-c}{b+c}\cot{(A/2)} gives \frac{B-C}{2}. Also, \frac{B-C}{2}=90 \deg - \frac{A}{2}, so that B and C can be evaluated. The third side is given by a=b \frac{\sin{A}}{\sin{B}}, or, a^{2}=b^{2}+c^{2}-2bc \cos{A}.
  • If two sides b and c and the angle B (opposite to side b) are given, then \sin{C}=\frac{c}{b}\sin{B}. And, A=180\deg -(B+C), and a=\frac{b \sin{A}}{\sin{B}} give the remaining elements.

By applying the cosine rule, we have:

\cos{B}=\frac{a^{2}+c^{2} - b^{2}}{2ac}, or if we manipulate this, we get

a^{2}-(2c\cos{B})a+(c^{2}-b^{2})=0

or, a=c \cos{B} \pm \sqrt{b^{2}-(c\sin{B})^{2}}

This equation leads to the following cases:

Case I:

If b<c\sin{B}, no such triangle is possible.

Case II:

Let b=c\sin{B}. There are further following two cases:

Sub-case II a:

B is an obtuse angle, that is, \cos{B} is negative. There exists no such triangle.

Sub-case II b:

B is an acute angle, that is, \cos {B} is positive. There exists only one such triangle.

Case III:

Let b >c \sin{B}. There are following two cases further here also:

Sub-case IIIa:

B is an acute angle, that is, \cos {B} is positive. In this case, two values of a will exist if and only if c\cos{B} > \sqrt{b^{2}-(c \sin{B})^{2}} or, c>b, which means two such triangles are possible. If c<b, only one such triangle is possible.

Sub-case IIIb:

B is an obtuse angle, that is, \cos{B} is negative. In this case, triangle will exist if and only if \sqrt{b^{2}-(c \sin{B})^{2}} > c |\cos{B}| \Longrightarrow b > c. So, in this case, only one such triangle is possible. If b <c, there exists no such triangle.

Note:

If one side a and angles B and C are given, then A=180 \deg -(B+C), and b=a \frac{\sin{B}}{\sin{A}} and c=a\frac{\sin{C}}{\sin{A}}.

If the three angles A, B and C are given, we can only find the ratios of the three sides a, b, and c by using the sine rule(since there are infinite number of similar triangles possible).

More theory later,

Nalin Pithwa

 

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