In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P ?

byAmlesh Mahato •June 09, 2021

0

Solution:

⇨ Given

⇨ PQ = 5 cm

⇨ PR + QR = 25 cm

⇨ PR = 25 - QR .....(i)

⇨ In ∆ PQR

⇨ From Pythagoras Theorem

$Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics$

$\large \Rightarrow PR^2= QR^2+PQ^2$

$\large \Rightarrow \left ( 25-QR \right )^2= QR^2+5^2$

$\large \Rightarrow 25^2-2\times 25\times QR+QR^2= QR^2+25$

$\large \Rightarrow 625-50QR+QR^2-QR^2-25=0$

$\large \Rightarrow 600=50QR$

$\large \Rightarrow QR=\frac{600}{50}$

$\large \Rightarrow QR=12$

∴Equation (i)

⇨ PR = 25 - 12

⇨ PR = 13 cm

∴ sin P = p/h = QR/PR

⇨ sin P = 12/13

∴ cos P = b/h = PQ/PR

⇨ cos P = 5/13

∴ tan P = p/b = QR/PQ

⇨ tan P = 12/5