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Napier's analogy: tan((B - C)/2) = ((b - c)/(b + c)) cot(A/2)

This type of question, asked 5 times: 9 May 2024 S1, 22 Apr 2025 S1, 26 Apr 2025 S1, 16 Apr 2026 S1, 15 May 2026 S1. All 5 questions below, oldest first.

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  1. Question 1

    9 May 2024 S1

    137
    Two sides of a triangle are 3+1\sqrt{3}+1 and 3−1\sqrt{3}-1 and the included angle is 60∘60^\circ, then the difference of the remaining angles is

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  2. Question 2

    22 Apr 2025 S1

    112
    In a triangle ABC with usual notations if, tan⁡(B−C2)=xcot A2\tan\left( \frac{B - C}{2} \right)=xcot\frac{\text{ }A}{2}, then x=x=

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  3. Question 3

    26 Apr 2025 S1

    120
    In a triangle ABC , with usual notations, cot⁡(A+B2)⋅tan⁡(A−B2)=\cot\left( \frac{A + B}{2} \right)\cdot \tan\left( \frac{A - B}{2} \right)=

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  4. Question 4

    16 Apr 2026 S1

    63
    In a triangle ABC, with usual notations a=3+1,b=3−1a = \sqrt{3} + 1, b = \sqrt{3} - 1 and ∠C=60∘\angle C = 60^\circ then the values of ∠A\angle A and ∠B\angle B respectively are

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  5. Question 5

    15 May 2026 S1

    53
    In △ABC\triangle ABC, if b=3+1b = \sqrt{3} + 1, c=3−1c = \sqrt{3} - 1 and A=π3A = \dfrac{\pi}{3}, then B−C=…B - C = \ldots

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