Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 years ago

□ show : d² = ab - st.?

△ABC.

bisector of ∠C intersects AB at P.

BC = a , AC = b , AP = s , BP = t , CP = d.

show : d² = ab - st.

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  • 3 years ago
    Favorite Answer

    •b/a=s/t

    bt = as.....(1)

    • (b²+d²-s²)/2bd = (d²+a²-t²)/2ad

    ...ab²+ad²-as² = bd² + ba² - bt²

    ...ad² - bd² = (ba²-ab²) + as.s - bt.t

    ...d²(a-b) = ab(a-b) +bt.s - as.t

    ...d²(a-b)= ab(a-b) - st(a-b)

    ...d² = ab - st

    🍭

    • Saurabh Dubey
      Lv 4
      3 years agoReport

      • cot(arccot 3 + arccot 7 + arccot 13 + arccot 21)

      • tan-¹[((1/3) + (1/7))/(1-(1/3)(1/7))] +
      tan-¹[(1/13)+(1/21)/(1-(1/21)(1/13)]
      = tan-¹(1/2) + tan-¹(1/8)
      = tan-¹[(1/2)+(1/8)/(1-(1/2)(1/8))]
      =tan-¹(10/15)
      =cot-1(15/10)
      cotcot-¹(15/10) = 15/10

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