SOLUTION
Given the following
[tex]a=1.9in,A=46.5^0,C=90^0[/tex]
Consider the image below
To solve the right triangle, we need to find the following:
[tex]b=\text{?,c}=\text{?and B=?}[/tex]
To find B, we use the sum of angles in a triangle
hence
[tex]\begin{gathered} A^0+B^0+C^0=180^0^{} \\ \text{where A=46.5}^0,C=90^0 \end{gathered}[/tex]
Substituting into the equation we have,
[tex]\begin{gathered} 46.5^0+B+90^0=180^0 \\ 136.5+B=180^0 \end{gathered}[/tex]
Subtract 136.5 from both sides
[tex]\begin{gathered} 136.5+B-136.5^0=180^0-136.5^0 \\ \text{Then} \\ B=180^0-136.5 \\ B=43.5^0 \end{gathered}[/tex]
hence
B = 43.5°
To find b, we use the trigonometry ratio for tangent
From the triangle above
[tex]\begin{gathered} \tan A=\frac{a}{b} \\ \text{Where A=46.5in, a=1.9in, b=?} \end{gathered}[/tex]
Substituting into the equation
[tex]\begin{gathered} \tan 46.5=\frac{1.9}{b} \\ \text{Then } \\ b=\frac{1.9}{\tan46.5} \\ \text{Where tan46.5=1.0538} \end{gathered}[/tex]
Then
[tex]b=1.8030[/tex]
hence
b = 1.8in to 1 decima place
To find c, we also apply trigonometry ratio for sine of an angle
[tex]\begin{gathered} \text{sinA}=\frac{opposite}{\text{hypotenuse}} \\ \text{Where } \\ A=46.5^0,\text{ opposite =1.9in},\text{ hypotenuse =c} \end{gathered}[/tex]
Substituting the values into the equation we have,
[tex]\begin{gathered} \sin 46.5=\frac{1.9}{c} \\ Multiply\text{ both sides by c, we have } \\ c\times\sin 46.5=\frac{1.9}{c}\times c \\ \text{Then } \\ c\times\sin 46.5=1.9 \end{gathered}[/tex]
Divide both sides by 1.9, we have
[tex]\begin{gathered} c=\frac{1.9}{\sin 46.5}=\frac{1.9}{0.7254} \\ \text{Then} \\ c=2.6193 \end{gathered}[/tex]
hence
c = 2.6 in
Therefore, to solve the right triangle, we have
Answer; B = 43.5°, b = 1.8in, c = 2.6 in
