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Mathematics - Triangles

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Published in: Mathematics
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Revision Notes on Solution of Triangles.

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  1. Revision Notes on Solution of Trianqles Sides of a Sine rule : angles opposite to them. sin A/ a sin B/ b triangle are proportional to the sine so, in AABC, 2A/abc. sin C/ c of This may also be written as (a/s in A) Cosine rule : In any AABC, (b/ sin B) (c/ sin C) cos A cos B cos C Trigonometric sin sin sin cos cos cos tan tan tan a2) / 2 bc 102) / 2 ac c2) / 2 ab ratios of half-angles : (s —b) (s —c) /bc] Al [ (s -c) (s -a) /ac] (s —a) (s —b) /ab] xls (s Xls (s NC(s a) / bc - b)/ ac c) / ab b) c) a) (s (s (s c) a) b) (s (s (s b)] Projection b c b c a rule : In any AABC, cos C + c cos B cos A + a cos C cos B + b cos A Area of a triangle If A denotes the area of the triangle ABC, then it can the be calculated in any of the following forms : 1/2 bc sin A = 1/2 ca sin B 1/2 ab sin C NS (s 1/2. 1/2. 1/2. c) (a2 sin B sin C)/ sin (B + (b2 (c2 sin C sin A)/ sin sin A sin B)/ sin
  2. Semi -perimeter of the triangle If S denotes the perimeter Napier' s analogy In any AABC , of the triangle ABC, then tan (B tan (C tan [ (A m-n theorem Consider divides (b (c B) c) cot A/ 2 cot B/ 2 cot c/ 2 a triangle ABC where D is a s such point m: n, on side BC the side the BC cot cot in the ratio following then as shown figure, Apollonius cot e cot e theorem m n In a triangle ABC, a B if n cot ß. m cot C. AD is the results median hold through A, AB2 + AC2 then If the three sides say a, b and c of a triangle are given, way . then that it in the good: angle A / 2bc. is obtained with the help of tan A/ 2 Angles B and C can also If two sides b and c and the tan (B (b c) This gives the value of the formula a) or cos A be obtained in the same included angle A are given, / (b + c) cot A/ 2 then using (B + Hence, both B and C can using the formula 900 - be evaluated. A/ 2 along Now, the 2bc cosA. with the last equation sides can be evaluated b sin A/s in B or a2 102 + C 2 If two sides b and c and the angle B (opposite to side we can easily obtain results, b) are given, the remaining then using elements sin C Some of the the following c/ b sinB, i mp ort ant 1800 b sin A/sinB (B + C) and b points which must be remembered include :
  3. 1. b < c sin B, there is no triangle possible (Fig. 1) 2. If b = c sin B and B is an acute angle, then only one triangle is possible (Fig. 2) 3. If c sin B < b < c and B is an acute angle, then there are two values of angle C (Fig. 3). 4. If c < b and B is an acute angle, then there is only one triangle (Fig. 4).

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