Question

in a triangle pqr angle Q is a right angle and QT is a perpendicular drawn on PR if PR is equal to 15 cm and q t is equal to 5 cm then the value of tan P + tanR upon tanR into 10 P is ​

Answer 1

in right angle Δ PQR at Q , PR = 15 & QT ⊥ PR  , QT = 5 , Then value of (TanP + TanR)/(TanR . TanP) = 3

Step-by-step explanation:

Area of Δ PQR

=(1/2) * PR * QT  = (1/2) * PQ * QR

=> PR * QT = PQ * QR

=> 15 * 5 = PQ * QR

=> PQ * QR = 75

Tan P = QR/PQ

Tan R = PQ/QR

(TanP + TanR)/(TanR . TanP)

= (QR/PQ  + PQ/QR) /( (PQ/QR) *(QR/PQ) )

=  ( QR² + PQ²)/ (PQ * QR)

= PR² / 75

= 15² / 75

= 3

(TanP + TanR)/(TanR . TanP) = 3