Given
Two triangles
Find
Length of the side and angle of triangle A1B1C1
Explanation
In triangle ABC
angle A = 90 degree
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B} \\ \sin B=\frac{3}{6}\times\sin A \\ \sin B=\frac{1}{2}\times\sin90\degree \\ B=30\degree \end{gathered}[/tex]
as we know , sum of all angles of triangle is 180 degree.
so,
[tex]\angle C=180\degree-90\degree-30\degree=60\degree[/tex]
now, from
[tex]\Delta A_1B_1C_1[/tex]
we have
[tex]\begin{gathered} A_1C_1=\frac{3}{1.5}=2 \\ B_1C_1=\frac{6}{1.5}=4 \\ A_1B_1=\sqrt{4^2-2^2}=\sqrt{16-4}=\sqrt{12}=2\sqrt{3} \end{gathered}[/tex]
Final Answer
Sides are
[tex]\begin{gathered} A_1C_1=2 \\ B_1C_1=4 \\ A_1B_1=2\sqrt{3} \end{gathered}[/tex]
angles are
[tex]\begin{gathered} \angle A=90\degree \\ \angle B=30\degree \\ \angle C=60\degree \end{gathered}[/tex]
