Question 25

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

Solution

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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