Question

In ∆PQR, ∠P = 90° . PQ = 24 cm, PR + QR = 32 cm, find 5sinR – 4secQ.

Answer 1

Answer:

I have gaven the answer

Answer 2

Answer:

in ΔPQR,PQ=24cm

               QR=26cm

               ∠QPR=90°

By using pythagoras theorem,

   ⇒QR²=QP²+PR²

   ⇒26²=24²+PR²

  ⇒676=576+PR²

  ⇒PR²=100

  ⇒PR=10cm---------------------------eq1

in ΔPKR,KR=8cm

              ∠PKR=90°

By using pythagoras theorem,

⇒PR²=PK²+KR²

⇒10²=PK²+8²

⇒100=PK²+64

⇒PK²=36

⇒PK=6cm