Solutions of Triangles

Solution of a right angled triangle:

(i) When two sides are given: Let the triangle be right angled at C. Then we can determine the remaining elements as given in the following table.

(ii) When a side and an acute angle are given: In this case, we can determine the remaining elements as given in the following table

Solution of a triangle in general:

(i)     When three sides a, b and c are given:

Δ = √s(s – a)(s – b)(s – c), where 2s = a + b + c = perimeter of triangle

(ii)   When two sides a, b and the included angle C are given:

and

(iii)  When one sides a and two angles A and B are given

A + B + C = 180º

⇒ C = 180º – A – B

b = ((a sin B)/(sin A)) and c = (a sin c)/(sin A) ⇒ Δ = (1/2) c a sin B

(iv)   When two sides a, b and the angle A opposite to one side is given

sin B = (b/a) sin A .....(i)  

c = 180º – (A + B), c = (a sin C)/(sin A)

Special Cases

Case I: When A is an acute angle:

(a)      If a < b sin A, there is no triangle. When a < b sin A, then (i), sin B > 1, which is impossible.

(b)      If a = b sin A, then only one triangle is possible which is right angled at B. When a = b sin A, then from sine rule. sin B = 1,  ∴ ∠B = 90º from fig. It is clear that CB = a = b sin A

Thus, in this case, only one triangle is possible which is right angled at B.

(c)      If a > b sin A, then three possibilities will arise:

(i)      a = b In this case, from sine rule sin B = sin A

 or B = 180º – A.

But B = 180º – A ⇒ A + B = 180º, which is not possible in a triangle. ∴ In this case, we get ∠A  = ∠B.

Hence, if b = a > b sin A then only one isosceles triangle ABC is possible in which ∠A  = ∠B.

(ii)   a > b In the following figure, Let AC = b, ∠CAX = A,  and a > b, also a > b sin A.

Now taking C as centre, if we draw an arc of radius a, it will intersect AX at one point B and hence only one ΔABC is constructed.

Also this arc will intersect XA produced at B' and ΔAB'C is also formed but this Δ is inadmissible (because ∠CAB' is an obtuse angle in this triangle)

Hence, if a > b and a > b sin A, then only one triangle is possible.

(iii) b > a (i.e., b > a > b sin A)

In fig. let AC = b, ∠CAX = A. Now taking C as centre, if we draw an arc of radius a, then it will intersect AX at two points B₁ and B₂. Hence if b > a > sin A, then there are two triangles.

Case II: When A is an obtuse angle:

► In this case, there is only one triangle, if a > b

Case III: b > c and B = 90º:

► Again the circle with A as centre and b as radius will cut the line only in one point. So, only one triangle is possible.

Case IV: b ≤ c and B = 90º

► The circle with A as centre and b as radius will not cut the line in any point. So, no triangle is possible.

► This is, sometimes called an ambiguous case.

Alternative Method:

► By applying cosine rule, we have cos B = (a² + c² – b²)/2ac

⇒ a² – (2c cos B)a + (c² – b²) = 0

⇒ a = cos B ± √(c cos B)² – (c² – b²)

⇒ a = c cos B  ± √b² – (c sin B)²

► This equation leads to following cases:

Case I:

► If b < sin B, no such triangle is possible.

Case II:

► Let b = c sin B, there are further following case.

(a) B is an obtuse angle

⇒ cos B is negative. There exists no such triangle.

(b)  B is an acute angle

⇒ cos B is positive. There exists only one such triangle.

Case III:

► Let b > c sin  B. There are further following cases :

(a)  B is an acute angle ⇒ cos B is positive.

► In this case two values of a will exists if and only if c cos B > √b² – (c sin B)² or c > b. Two such triangle is possible. If c < b, only one such triangle is possible.

(b)  B is an obtuse angle ⇒ cos B is negative. In this case triangle will exist if and only if √(b² – (c sin B)²) > |c cos B| ⇒ b > c. So, in this case only one such triangle is possible. If b < c there exists no such triangle.