## 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, 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, 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 , 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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