Question

In ΔPQR, PT ⊥ QR, PQ = 10 cm, QT = 6 cm, TR = 15 cm. Find the value of sin x + tan y.

Answer 1

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`