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JEE Main Mock Test Series 2020 & Previous Year Papers

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  • Sine rule: Sides of a triangle are proportional to the sine of the angles opposite to them. So, in ΔABC,  
    sin A/a = sin B/b = sin C/c = 2Δ/abc.
    This may also be written as (a/sin A) = (b/sin B) = (c/sin C)
  • Cosine rule: In any ΔABC,
    cos A = (b2 + c2 – a2) /2bc
    cos B = (a2 + c2 – b2)/2ac
    cos C = (a2 + b2 - c2)/2ab
  • Trigonometric ratios of half-angles: 
    sin A/2 = √[(s-b)(s-c)/bc]
    sin B/2 = √[(s-c) (s-a)/ac]
    sin C/2 = √[(s-a) (s-b)/ab]
    cos A/2 = √s(s - a)/bc
    cos B/2 = √s(s - b)/ac
    cos C/2 = √s(s - c)/ab
    tan A/2 = √[(s - b) (s - c)/s(s - a)]
    tan B/2 = √[(s - c) (s - a)/s(s - b)]
    tan C/2 = √[(s - a) (s - b)/s(s - c)]
  • Projection rule: In any ΔABC, 
    a = b cos C + c cos B
    b = c cos A + a cos C
    c = a cos B + b cos A
  • Area of a triangle
    If Δ denotes the area of the triangle ABC, then it can be calculated in any of the following forms:
    Δ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin C
    Δ = √s(s - a)(s – b)(s - c)
    Δ = 1/2. (a2 sin B sin C)/ sin(B + C)
    = 1/2. (b2 sin C sin A)/ sin (C + A)
    = 1/2. (c2 sin A sin B)/ sin (A + B)
  • Semi-perimeter of the triangle
    If S denotes the perimeter of the triangle ABC, then s = (a + b + c)/2
  • Napier’s analogy 
    In any ΔABC,
    tan [(B – C)/2] = (b – c)/(b + c) cot A/2
    tan [(C – A) /2] = (c – a)/(c + a) cot B/2
    tan [(A – B) /2] = (a – b)/(a + b) cot C/2
  • m-n theorem
    Consider a triangle ABC where D is a point on side BC such that it divides the side BC in the ratio m: n, then as shown in the figure, the following results hold good:
     Notes | EduRev(m + n) cot θ = m cot α – n cot ß.
    (m + n) cot θ = n cot B – m cot C.
  • Apollonius theorem
    In a triangle ABC, if AD is the median through A, then
    AB+ AC2 = 2(AD2 + BD2).
  • If the three sides say a, b and c of a triangle are given, then angle A is obtained with the help of the formula
    tan A/2 = √(s - b) (s - c) / s(s - a) or cos A = b2 + c2 - a2 / 2bc.
    Angles B and C can also be obtained in the same way.
  • If two sides b and c and the included angle A are given, then
    tan (B – C) /2 = (b – c)/ (b + c) cot A/2
    This gives the value of (B- C)/2.
    Hence, using (B + C)/2 = 90o - A/2 along with the last equation both B and C can be evaluated. Now, the sides can be evaluated using the formula
    a = b sin A/sin B or a2 = b2 + c2 – 2bc cosA.
  • If two sides b and c and the angle B (opposite to side b) are given, then using the following results, we can easily obtain the remaining elements
    sin C = c/b sinB, A = 180– (B + C) and b = b sin A/sinB
  • Some of the important points which must be remembered include:
    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). Notes | EduRev
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