Sum

In the given Fig, `angle` R is the right angle of `triangle` PQR. Write the following ratios.

(i) sin P (ii) cos Q (iii) tan P (iv) tan Q

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#### Solution

(i) Sin P = `" Opposite side of ∠P"/" Hypotenuse "` = `["QR"]/["PQ"]`

(ii) CosQ = `"Adjacent side of ∠Q"/" Hypotenuse "` = `["QR"]/["PQ"]`

(iii) tan P = `" Opposite side of ∠P"/"Adjacent side of ∠P"` = `["QR"]/["PR"]`

(iv) tan Q = `" Opposite side of ∠P"/"Adjacent side of ∠P"` = `["PR "]/["QR"]`

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