Question

In ∆PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.

Answer 1

We are given the following information in the form of the triangle

To find ∠P and ∠R

Now in ΔPQR

`cos P = (PQ)/(PR)`

`cos P = 3/6`  .....(1)

`= 1/2 `

Now we know that

`cos 60^@ = 1/2` ....(2)

Now by comparing equation (1) and (2)

We get,

`∠P = 60^@ ....(3)`

Now we have

`sin P =(QR)/(PR)`

`sin 60^@= (QR)/6`

Now we know that

`sin 60^@ = sqrt3/2`

Therefore,

`sqrt3/2 = (QR)/6`

Now by cross multiplying

We get

`6 xx sqrt3 = 2 xx QR`

`=> 6sqrt3 = 2QR`

`=> QR = (6sqrt3)/2`

`=> QR = 3sqrt3`

Therefore

`QR = 3sqrt3 cm` .....(4)

Now we know that

`cos R = (QR)/(PR)`

`cos R = (3sqrt3)/2`

`=> cos R = sqrt3/2`  ....(5)

Now we know,

`cos 30^@ = sqrt3/2` ....(6)

Now by comparing equation (5) and (6)

We get,

∠R = 30° ...(7)

Hence from equation (3) and (7)

`∠P = 60^@ and ∠R = 30^@`