Given:
In the given triangle the measure of two angles are 102 degree and 28 degrees and the sides of the triangle are a, 27 cm, c.
To find:
The value of a.
Solution:
Let the given triangle be ABC, such that,
[tex]m\angle B=28^\circ[/tex]
[tex]m\angle C=102^\circ[/tex]
[tex]b=27\ cm[/tex]
Using angle sum property of triangles, we get
[tex]m\angle A+m\angle B+m\angle C=180^\circ[/tex]
[tex]m\angle A+28^\circ+102^\circ=180^\circ[/tex]
[tex]m\angle A=180^\circ-28^\circ-102^\circ[/tex]
[tex]m\angle A=50^\circ[/tex]
According to the Law of Sines:
[tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}[/tex]
[tex]\dfrac{\sin 50^\circ }{a}=\dfrac{\sin 28^\circ }{27}[/tex]
[tex]\dfrac{27\times \sin 50^\circ }{\sin 28^\circ}=a[/tex]
[tex]\dfrac{27\times \sin 50^\circ }{\sin 28^\circ}=a[/tex]
[tex]a\approx 44.06[/tex]
Therefore, the value of a is about 44.06 cm.
