Please solve question no.7,8 ,12,13 and10

$$\begin{gathered} Dear student, \\ 7) \\ In △ABC \\ AB=6 \\ BC=10 \\ also from pythagores theorem, \\ A{B}^2+A{C}^2=B{C}^2 \\ 36+A{C}^2=100 \\ A{C}^2=100-36=64 \\ AC=8 \\ From△ABC \\ \sin B=\frac{AC}{BC}=\frac{8}{10}=\frac{4}{5} \\ \sin B=\frac{4}{5} \\ \cos B=\frac{AB}{BC}=\frac{6}{10}=\frac{3}{5} \\ \cos B=\frac{3}{5} \\ \sin C=\frac{AB}{BC}=\frac{6}{10}=\frac{3}{5} \\ \sin C=\frac{3}{5} \\ \cos C=\frac{AC}{BC}=\frac{8}{10}=\frac{4}{5} \\ \sin B \cos C+\sin C \cos B=\frac{4}{5}\times \frac{4}{5}+\frac{3}{5}\times \frac{3}{5} \\ =\frac{16}{25}+\frac{9}{25} \\ =\frac{25}{25} \\ =1 \\ 12) \tan A=\frac{x}{y} \\ from trigonometric identiy \\ 1+{\tan }^{2}A=se{c}^{2}A \\ 1+\frac{x^2}{y^2}=se{c}^{2}A \\ \frac{y^2+x^2}{y^2}=se{c}^{2}A \\ secA=\frac{\sqrt{x^2+y^2}}{y} \\ also \\ \cos A=\frac{1}{secA} \\ \cos A=\frac{y}{\sqrt{x^2+y^2}} \end{gathered}$$

The solution/s to your 7th and 12th  query has/have been provided above
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