The given triangle ACB is a right angle triangle with 90 degree at angle C
From the figure we have;
Angle C = 90
CB = 11.9
AC = 10
We need to find the value of angle A and the side AB
For angle Ө;
The side Adjacent to angle Ө is AC and the opposite side to angle Ө is CB
Thus., From the trigonometric ratio;
The ratio of the Opposite side to the adjacent side the tangent of the angle.
[tex]\tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}}[/tex]
Substitute the value, Opposite side BC = 11.9 and Adjacent side AC = 10
[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan \theta=\frac{BC}{AC} \\ \tan \theta=\frac{11.9}{10} \\ \tan \theta=1.19 \\ \theta=\tan ^{-1}(1.19) \\ \theta=49.95^o \end{gathered}[/tex]
Thus, the missing angle is 49.95°
Now, for the side AB;
Apply the trignometric ratio of sin of angle 49.95°
The ratio of the adjacent side to the hypotenuse is the sine of the angle.
[tex]Sin\theta=\frac{Adjacent\text{ Side}}{\text{Hypotenuse}}[/tex]
Substitute the value; Adjacent side AC = 10 and Hypotenuse AB and sin 49.95° = 0.765
[tex]Sin\theta=\frac{Adjacent\text{ Side}}{\text{Hypotenuse}}[/tex]
