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| Solution of triangles |
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| We have studied that a triangle has six parts or six elements viz three sides and three angles. From geometry, we know that when any three elements are given of which necessarily a side is given, the triangle is completely determined i.e, remaining three elements can be determined. The process of determining the unknown elements knowing the known elements is known as the solution of a triangle. In practice, there are four different cases for which the solution is discussed as under. |
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| Case 1: When all three sides are given. |
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| To solve a triangle given the three sides a, b, c. |
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| To determine angles A, B and C. |
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| The angles A, B and C are determined by using the following relations: |
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| Example: |
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| The sides of a triangle are 20, 30, and 21. Find the greatest angle. |
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| Suggested answer: |
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| The greatest angle is opposite to the side whose length is 30. |
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| Let a = 20, b = 30, c = 21. |
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| We have to find angle B. |
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| Case 2: |
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| When two sides and the angle included between these are given. |
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| Example: |
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| If b = 251, C = 147, A = 47o, find the remaining angles. |
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| (Use Napier's rule) |
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| Suggested answer: |
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| = 0.268 x 2.2998 |
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| Case 3: |
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| To solve a triangle having given two angles and a side. |
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| Let the given parts be denoted by B, C, a, then the third angle A can be found from the relation. |
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| Similarly, c can be found from the equation. |
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| log c = log a + log sin C - log sin A |
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| Example: |
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| Solve the triangle ABC, given a =18, A = 25o, B = 108o. |
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| Suggested answer: |
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| C = 180o - (25 + 108o) = 47o |
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| Taking log on both sides, we get |
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| log b = log a + log (sin B) - log (sin A) |
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| = log 18 + log (sin 180o) - log (25o) |
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| = log 18 + log (sin 72o) - log (25o) |
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| = log 18 + log (0.9511) - log (0.4226) |
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| (from Trigonometric table) |
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| = 1.2553 - 1 + 0.9782 + 1 - 0.6295 |
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| log b = 1.6076 |
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| Taking antilog, we get b = 40.52. |
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| Similarly using |
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| log c = log a + log (sin C) - log (sin A), |
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| we can evaluate the value of C as 31.23o. |
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| Case 4: |
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| When two sides b, c and an angle B opposite to one of the given sides are given. |
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| log sin C = log c + log B - log b which determines C. |
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| The equation (i) in general leads to two solutions say C1 and C2 (say such that C1 + C2 = 180), suppose C1 is acute. |
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| Corresponding to C1 and C2, we have two values a1, a2 (say of the remaining side a). They are given by |
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| From equation (i), if an acute angle C1 is a solution of (i) then |
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| 180o - C1 is also a solution and there is an ambiguity. However if |
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| From geometry, C must be acute. For if C is 90o or more than B then C must be greater than 90o. This is impossible, since a triangle can have only one right angle or more in size. |
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| If b < c, there may be two triangles having the given elements b, c and B. One having an acute angle C1 and second having an angle |
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| Fig (a) Fig (b) |
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| Fig (c) |
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| Figures (a) and (b) show its two triangles. |
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| Figure (c) shows two triangles superimposed and suggests a geometric method of drawing the two triangles. |
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| From figure (c), it is evident that if b < c sinB, there is no triangle. |
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| The following rules may be used to solve a triangle with given b, c and B. |
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| Step 1: |
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| Use (i) to find log (sin C). |
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| Step 2: |
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| If log (sin C) comes out as positive, sin C would have to be greater than 1 and there is no solution. |
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| Step 3: |
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| If log (sin C) comes out as negative or zero, find the corresponding acute angle C. |
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Then, find A = 180o - (B + C) and finally get 'a' by using the law of sines (this completes the solution of  ). |
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| Step 4: |
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| If bo - C instead of C. |
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| Example: |
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| If in a DABC, a = 97, b = 119, A = 50o, find B and c given that: |
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log (sin (70o, 1'))  9.9730318 . |
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| Suggested answer: |
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| \ B = 70o0'57" or (180o - 70o0'57") i.e., 109o59'03" |
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| Since a < b, both the values of B is admissible and then case is ambiguous. |
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