Question

In ∆PQR, Q = 90°, PR = 10√2 cm and QR = 10cm. Find the values of sec P,cot P and cosec P.​

Answer 1

in triangle PQR

by using pgt theorem

h^ = p^ + b^

10√2^ = 10^ + PQ^

100×2 = 100 + PQ^

200 = 100 + PQ^

200- 100 = PQ^

100 = PQ^

√100 = PQ

10cm = PQ

sec P = H/B = 10√2/ 10

sec P = √2

cot P B/ P = 10/ 10 = 1

cosecP = H/ P = 10√2/ 10

cosecP = √2

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