Question

in ∆PQR angle Q is 90 degree and PQ=4cm and PR=9cm. Find :- sin P, tan R, sin R+cos R. I will mark the correct answer the brainliest​

Answer 1

Answer:

QR=√65

sinP=QR/PR= √65/9

tanR=PQ/QR=4/√65

sinR+cosR = (PQ+QR)/PR = (4+√65)/9

Answer 2

Answer:

sin P=

$\frac{ \sqrt{65} }{4}$

tan R

$\frac{4}{ \sqrt{65} }$

sirR+cosR

$\frac{4 + \sqrt{65} }{9}$

Step-by-step explanation:

since Q is a right angle triangle PR is the hypotenuse.

by using

${pr }^{2} = {pq}^{2} + {qr}^{2}$

we can find QR. then substitute the formulas of SIN ,TAN AND COS