Question

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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] = 12k Read 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:

Step-by-step explanation:

Given that,

sin θ = 5/13 = PQ/PR

By using Pythagoras theorem,

PR² = PQ² + QR²

( 13 )² = ( 5 )² + QR²

169 = 25 + QR²

QR² = 169 - 25

QR² = 144

QR = √144

QR = 12

∴ cos θ = QR/PR = 12/13

And tan θ = PQ/ QR = 5/12