1) The values are [tex]Q \approx 42.969^{\circ}[/tex], [tex]P \approx 52.032^{\circ}[/tex] and [tex]QR \approx 15.036[/tex], respectively.
2) The values are [tex]C = 22^{\circ}[/tex], [tex]DC = 10.429[/tex] and [tex]BC = 21.016[/tex], respectively.
How to find missing sides and angles by law of sines and law of cosines
In this exercise we shall apply concepts of law of sines and law of cosines corresponding to each case:
1) Now we proceed to find the length of QR by law of cosines:
[tex]QP = \sqrt{PR^{2}+QR^{2}-2\cdot PR\cdot QR \cdot \cos R}[/tex]
[tex]19^{2} = 13^{2}+QR^{2}-2\cdot 13\cdot QR\cdot \cos 85^{\circ}[/tex]
[tex]QR^{2}-2.266\cdot QR -192 = 0[/tex] (1)
There are two solutions: [tex]QR_{1} \approx 15.036[/tex] and [tex]QR_{2}\approx -12.770[/tex], of which the former one is reasonable.
Then, we obtain the missing angles by law of sine:
[tex]\frac{QP}{\sin R} = \frac{PR}{\sin Q} = \frac{QR}{\sin P}[/tex]
[tex]\frac{19}{\sin 85^{\circ}} = \frac{13}{\sin Q} = \frac{15.036}{\sin P}[/tex] (2)
Whose solutions are: [tex]Q \approx 42.969^{\circ}[/tex], [tex]P \approx 52.032^{\circ}[/tex].
The values are [tex]Q \approx 42.969^{\circ}[/tex], [tex]P \approx 52.032^{\circ}[/tex] and [tex]QR \approx 15.036[/tex], respectively. [tex]\blacksquare[/tex]
2) The lengths of [tex]DC[/tex] and [tex]BC[/tex] is found by law of sines:
[tex]\frac{BD}{\sin C} = \frac{DC}{\sin B} = \frac{BC}{\sin D}[/tex]
[tex]\frac{12}{\sin 22^{\circ}} = \frac{DC}{\sin 19^{\circ}} = \frac{BC}{\sin 139^{\circ}}[/tex] (3)
Whose solutions are [tex]DC = 10.429[/tex] and [tex]BC = 21.016[/tex].
The values are [tex]C = 22^{\circ}[/tex], [tex]DC = 10.429[/tex] and [tex]BC = 21.016[/tex], respectively. [tex]\blacksquare[/tex]
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