Answer:
[tex]A=60^0, B=30^0, C=90^0\\a=3.46, b=2, c=4[/tex]
Step-by-step explanation:
In the diagram below:
First, we determine the value of [tex]\beta[/tex]
[tex]\alpha+\beta=90^0 $ (Other Angles of a Right Triangle)$\\60+\beta=90^0\\\beta=90^0-60^0=30^0[/tex]
To determine the value of side a, we apply the Sine rule
[tex]\dfrac{c}{Sin C} =\dfrac{a}{Sin \alpha} \\\dfrac{4}{Sin 90}=\dfrac{a}{Sin 60}\\ a=\dfrac{4*sin60}{sin 90}\\a=3.46[/tex]
Similarly, to determine the value of side b, we apply the Sine rule
[tex]\dfrac{c}{Sin C} =\dfrac{b}{Sin \beta} \\\dfrac{4}{Sin 90}=\dfrac{b}{Sin 30}\\ b=\dfrac{4*sin30}{sin 90}\\b=2[/tex]
Therefore:
[tex]A=60^0, B=30^0, C=90^0\\a=3.46, b=2, c=4[/tex]
