1. $$\triangle$$PQR is right angled at Q. If secP = 13/5, then what is the value of sinR ?





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  • By: anil on 05 May 2019 03.28 am
    Given : $$sec P$$ = $$frac{13}{5}$$ Also, $$sec P=frac{PR}{PQ}=frac{13}{5}$$ Let PR = 13 cm and PQ = 5 cm To find : $$sin R=frac{PQ}{PR}$$ = $$frac{5}{13}$$ => Ans - (A)
  • By: anil on 05 May 2019 03.28 am
    Given : $$sec P$$ = $$frac{13}{5}$$ Also, $$sec P=frac{PR}{PQ}=frac{13}{5}$$ Let PR = 13 cm and PQ = 5 cm To find : $$sin R=frac{PQ}{PR}$$ = $$frac{5}{13}$$ => Ans - (A)
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