The Solution:
Given:
Part(a)
(i) The value of the ratio of

Applying the Law of sine:

In this case,
![\begin{gathered} \sin\angle PQR=\sin\angle Q \\ \sin\operatorname{\angle}PRQ=\sin\operatorname{\angle}R \\ \sin\operatorname{\angle}RPQ=\sin\operatorname{\angle}P \end{gathered}]()
So, the ratio
![\frac{\sin\operatorname{\angle}PRQ}{\sin\operatorname{\angle}RPQ}=\frac{\sin\operatorname{\angle}R}{\sin\operatorname{\angle}P}]()
Applying the Law of sine to get the above ratio, we get

Substituting, we get
![\frac{\sin\operatorname{\angle}R}{2x}=(\sin\angle P)/(3x)]()
Cross multiplying, we get
![\begin{gathered} (\sin\angle R)/(\sin\angle P)=(2x)/(3x)=(2)/(3) \\ Thus, \\ \frac{\sin\operatorname{\angle}PRQ}{\sin\operatorname{\angle}RPQ}=\frac{\sin\operatorname{\angle}R}{\sin\operatorname{\angle}P}=(2)/(3) \end{gathered}]()
(ii) To find:

Applying the Law of Cosine:

Where,

Substituting, we get

Thus,

Part (b)
To find the value of x if the area of the triangle PQR is 12 square centimeters.
By area of triangle formula:

Substituting these values, we get

Dividing both sides by 5.8094, we get

Thus, the value of x is 2.0325