Question

Find the values of the six trigonometric functions for angle Ѳ, when PQ=48 and QR=64​

Answer 1

Answer:

Here's what I get    

Step-by-step explanation:

∆PQR is a right triangle.

PR² = PQ² + QR² = 48² + 64² = 2304 + 4096 = 6400

PR = √6400 = 80  

PQ:QR:PR = 48:64:80 = 3:4:5

We can consider ∆PQR as a 3:4:5 triangle.

$\sin \theta = \dfrac{4}{5} \\\\\cos \theta = \dfrac{3}{5}\\\\\tan \theta = \dfrac{4}{3}\\\\\cot \theta = \dfrac{3}{4} \\\\\csc \theta = \dfrac{5}{4}\\\\\sec \theta = \dfrac{5}{3}$

Answer 2

Answer:

The formula used for Pythagoras Theorem.

(Hypotenuse)² = (Base)² + (Perpendicular)²

We have

PQ = 48, QR = 64 and PR = ?

⇒ (PR)² = (PQ)² + (QR)²

⇒ (PR)² = (48)² + (64)²

⇒ (PR)² = 2304 - 4096 = 64 00

⇒ PR = 80

The six trigonometric functions we have are:

• sine = sin θ = Perpendicular ÷ hypotenuse  = PQ ÷ PR = 48 ÷ 80 = 0.6

• cosine = cos θ = Base ÷ hypotenuse   = QR ÷ PR = 64 ÷ 80 = 0.8

• tangent = tan θ = Perpendicular ÷ Base  = PQ ÷ QR = 48 ÷ 64 = 0.75

• cosecant = cosec θ = hypotenuse ÷ Perpendicular  = PR ÷ PQ = 80 ÷ 48 = 1.67

• secant = sec θ = hypotenuse ÷ Base  = PR ÷ QR = 80 ÷ 64 = 1.25

• cotangent = cot θ = Base ÷ Perpendicular  = QR ÷ PQ = 64 ÷ 48 = 1.34