Answer:
[tex]\angle PRQ = 50^\circ[/tex]
PR = 5.74 units and
PQ = 8.53 units
Step-by-step explanation:
We are given the following details:
[tex]\angle RPQ =99^\circ[/tex]
[tex]\angle PQR =31^\circ\\\text{Side QR} = 11\ units[/tex]
To find:
[tex]\angle PRQ =?\\Side\ PR = ?\\Side\ PQ = ?[/tex]
We know that the sum of all the angles of a triangle is equal to [tex]180^\circ[/tex]
i.e.
[tex]\angle PQR +\angle PRQ +\angle RPQ =180^\circ\\\Rightarrow 31^\circ +\angle PRQ +99^\circ =180^\circ\\\Rightarrow \angle PRQ = 180-130\\\Rightarrow \angle PRQ = 50^\circ[/tex]
To find the sides, we can use Sine rule:
As per Sine rule:
[tex]\dfrac{a}{sin A} =\dfrac{b}{sin B} =\dfrac{c}{sin C}[/tex]
Where a, b and c are the sides opposite to [tex]\angle A,\angle B,\angle C[/tex] respectively.
Using Sine rule in given triangle:
[tex]\dfrac{11}{sin 99} =\dfrac{b}{sin 31} =\dfrac{c}{sin 50}\\\\\text{Solving }\dfrac{11}{sin 99} =\dfrac{b}{sin 31} \\\Rightarrow b = \dfrac{11}{sin 99} \times sin31\\\Rightarrow b =5.74\ units\\\\\text{Now, Solving }\dfrac{11}{sin 99} =\dfrac{c}{sin 50} \\\Rightarrow c = \dfrac{11}{sin 99} \times sin50\\\Rightarrow c =8.53\ units[/tex]
So, sides are
PR = 5.74 units and
PQ = 8.53 units
