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Sagot :
Given:
[tex]A=15^\circ, C=120^\circ,b=3[/tex].
To find:
The length of side c.
Solution:
According to the angle sum of property of a triangle, the sum of all interior angles of a triangle is 180 degrees.
[tex]m\angle A+m\angle B+m\angle C=180^\circ[/tex]
[tex]15^\circ+m\angle B+120^\circ=180^\circ[/tex]
[tex]m\angle B+135^\circ=180^\circ[/tex]
[tex]m\angle B=180^\circ-135^\circ[/tex]
[tex]m\angle B=45^\circ[/tex]
According to the Law of sines,
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
Using Law of sines, we get
[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
[tex]\dfrac{3}{\sin 45^\circ}=\dfrac{c}{\sin 120^\circ}[/tex]
[tex]\dfrac{3}{\dfrac{1}{\sqrt{2}}}=\dfrac{c}{\dfrac{\sqrt{3}}{2}}[/tex]
[tex]3\sqrt{2}\times \dfrac{\sqrt{3}}{2}=c[/tex]
On further simplification, we get
[tex]\dfrac{3\sqrt{6}}{2}=c[/tex]
[tex]3.674235=c[/tex]
[tex]c\approx 3.7[/tex]
Therefore, the length of side c is about 3.7 units.
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