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प्रश्न
In ΔPQR, PT ⊥ QR, PQ = 10 cm, QT = 6 cm, TR = 15 cm. Find the value of sin x + tan y.
बेरीज
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उत्तर
Given:
PT ⊥ QR, PQ = 10 cm, QT = 6 cm, TR = 15 cm
Step 1:
ΔPQT,
PT = `sqrt(PQ^2 − QT^2)`
PT = `sqrt(10^2 − 6^2)`
PT = `sqrt64`
PT = 8 cm
Step 2:
sin x ...[angle at P in ΔPQR]
sin x = `"opposit"/"hypotenuse" = (QT)/(PQ)`
sin x = `6/10`
sin x = `3/5`
Step 3:
tan y ...[angle at R in right △PRT]
tan y = `"opposite"/"adjacent"`
= `(PT)/(TR) `
= `8/15`
= `sin x + tan y` ...[Adding]
= `3/5 + 8/15`
= `9/15 + 8/15`
= `17/15`
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