Geometry help....?

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  • Indica
    Lv 7
    9 months ago
    Favorite Answer

    ∠BPA=∠C, ∠BAP=π/2−B hence ∠ABP=π/2+B−C

    Sine Rule to ΔABP : 4/sin(π/2+B−C) = 2R = 4√2 ⟹ cos(B−C) = 1/√2 ⟹ sec(B−C) = √2

    • CK9 months agoReport

      Thank you sooo much sir again and again! :))

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  • 9 months ago

    Let altitude AD of acute angled triangle ABC be produced

    which meets its circumcircle at P such that AP = 4.

    If circumradius of triangle is 2 root w then find the value of sec (B - C)

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