Question

∆PQR is a right angled at Q PQ = 4 cm and PR = 8 cm find QR sin R tan R







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Answer 1

Answer:

Given: △PQR is right-angled at Q. PQ = 4 cm RQ = 3 cm Required to find: sin P, sin R, sec P, sec R Given △PQR, By using Pythagoras theorem to △PQR, we get PR2 = PQ2 + RQ2 Substituting the respective values, PR2 = 42 + 32 PR2 = 16 + 9 PR2 = 25 PR = √25 PR = 5 ⇒ Hypotenuse = 5 By definition, sin P = PerpendicularsideoppositetoanglePHypotenusePerpendicularsideoppositetoanglePHypotenuse sin P = RQPRRQPR ⇒ sin P = 3535 And, sin R = PerpendicularsideoppositetoangleRHypotenusePerpendicularsideoppositetoangleRHypotenuse sin R = PQPRPQPR ⇒ sin R = 4545 And, sec P = 1cosP1cosP sec P = HypotenuseBasesideadjacentto∠PHypotenuseBasesideadjacentto∠P sec P = PRPQPRPQ ⇒ sec P = 5454 Now, sec R = 1cosR1cosR sec R = HypotenuseBasesideadjacentto∠RHypotenuseBasesideadjacentto∠R sec R = PRRQPRRQ ⇒ sec R = 53

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Answer 2

The answer is given above in the picture....