Question

P.
3) In figure, Q=90°,
R=0, PQ=12,
QR = 5, PR = 13.
Find the value of :
cos Ꮎ .​

Answer 1

Answer:

i. Take the given trigonometric ratio as 13k equation (i). sin θ = 5/13 .. .(i) [Given] By using the definition write the trigonometric ratio of sin θ and take it as equation (ii). In right angled ∆PQR, ∠R = θ Let the common multiple be k. ∴ PQ = 5k and PR = 13k Find QR by using Pythagoras theorem. PR2 = PQ2 + QR2 … [Pythagoras theorem] ∴ (13k)2 = (5k)2 + QR2 ∴ 169k2 = 25k2 + QR2 ∴ QR2 = 169k2 – 25k2 = 144k2 ∴ QR = √(144k2) . . . [Taking square root of both sides] = 12kRead more on Sarthaks.com - https://www.sarthaks.com/850966/in-right-angled-pqr-q-90-r-and-if-sin-5-13-then-find-cos-and-tan

Answer 2

Answer:

i. PQ = 12, QR = 5 [Given] In APQR, ∠Q = 90° [Given] ∴ PR2 = QR2 + PQ2 [Pythagoras theorem] = 25 + 144 ∴ PR2 = 169 ∴ PR = 13 units [Taking square root of both sides] ii. In right angled APQR, seg QS is the median on hypotenuse PR. ∴ QS = (1/2) PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse] = (1/2) x 13 ∴ l(QS) = 6.5 unitsRead more on Sarthaks.com - https://www.sarthaks.com/849538/in-pqr-q-90-pq-12-qr-5-and-qs-is-a-median-find-l-qs