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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