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Sides given: find the area or a half-angle ratio using s and Heron's formula

This type of question, asked 8 times: 19 Apr 2025 S1, 20 Apr 2025 S2, 13 Apr 2026 S1, 17 Apr 2026 S1, 18 Apr 2026 S1, 19 Apr 2026 S1, 19 Apr 2026 S2, 12 May 2026 S2. All 8 questions below, oldest first.

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  1. Question 1

    19 Apr 2025 S1

    122
    The ratios of sides in a triangle ABC are 5:12:135:12:13 and its area is 270 . Then sides of the triangle are

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  2. Question 2

    20 Apr 2025 S2

    120
    In a triangle ABC with usual notations if a=13a = 13, b=14,c=15b= 14,c= 15 Then sin⁡A=\sin A=

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  3. Question 3

    13 Apr 2026 S1

    63
    In a triangle ABC, with usual notations if s−a11=s−b12=s−c13\dfrac{s - a}{11} = \dfrac{s - b}{12} = \dfrac{s - c}{13} and λtan⁡2A2=455\lambda\tan^2\dfrac{A}{2} = 455, then λ=\lambda =

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  4. Question 4

    17 Apr 2026 S1

    64
    In △ABC\triangle ABC, if a=13,b=14,c=15a = 13, b = 14, c = 15, then the value of sin⁡A+cos⁡A\sin A + \cos A is...

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  5. Question 5

    18 Apr 2026 S1

    64
    In △ABC\triangle ABC, with usual notation, if a=13,b=14,c=15a = 13, b = 14, c = 15, then the sum of the values of sin⁡(A2)\sin\left(\dfrac{A}{2}\right) and sin⁡A\sin A is....

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  6. Question 6

    19 Apr 2026 S1

    65
    In △ABC\triangle ABC, with usual notations, if the sides aa, bb and cc are in the ratio 18:17:718 : 17 : 7, then cot⁡A2:cot⁡B2:cot⁡C2=\cot\dfrac{A}{2} : \cot\dfrac{B}{2} : \cot\dfrac{C}{2} =

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  7. Question 7

    19 Apr 2026 S2

    64
    If ABC is a triangle of area Δ\Delta with a=2,b=72,c=52a = 2, b = \dfrac{7}{2}, c = \dfrac{5}{2}, where a,b,ca, b, c are the lengths of the sides of the triangle opposite to angles A, B and C respectively, then 2sin⁡A−sin⁡2A2sin⁡A+sin⁡2A\dfrac{2\sin A - \sin 2A}{2\sin A + \sin 2A} is equal to...

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  8. Question 8

    12 May 2026 S2

    55
    In △ABC\triangle ABC, if l(AB)=8,l(BC)=5,l(AC)=7l(AB) = 8, l(BC) = 5, l(AC) = 7, then cot⁡A2+cot⁡B2+cot⁡C2\cot\dfrac{A}{2} + \cot\dfrac{B}{2} + \cot\dfrac{C}{2} is equal to......

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