Question

IN TRIANGLE PQR RIGHT ANGLED AT Q PR+QR=25 CM AND PQ=5 CM. DETERMINE THE VALUE OF SIN P, COS P AND TAN P​

Answer 1

PR+QR=25 CM

PR=25-QR

PQ=5 CM

${h}^{2} = {p}^{2} + {b}^{2} \\ \sin( p) = ( \frac{p}{h} ) = \frac{12}{15} \\ cos \:( p) = \frac{5}{13} \\ \tan(p) = \frac{12}{5}$

Answer 2

Step-by-step explanation:

Given PR + QR = 25 , PQ = 5

PR be x.  and QR = 25 - x 

Pythagoras theorem ,PR2 = PQ2 + QR2

x2 = (5)2 + (25 - x)2

x2 = 25 + 625 + x2 - 50x

50x = 650

x = 13

 

 PR = 13 cm

QR = (25 - 13) cm = 12 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5