Answer:
49.3 m
Step-by-step explanation:
To find the perimeter of the given right triangle, use trigonometric ratios to determine the lengths of the remaining two sides, and then sum them together.
To find the length of the side opposite the given angle, we can use the tangent ratio:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=\dfrac{O}{A}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$O$ is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\end{array}}[/tex]
In this case:
Substitute the values into the tangent ratio and solve for O:
[tex]\tan 42^{\circ}=\dfrac{O}{15.2}\\\\\\O=15.2\tan 42^{\circ}\\\\\\O=13.686141473327...[/tex]
Now, use the cosine trigonometric ratio to find the length of the hypotenuse:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Cosine trigonometric ratio}}\\\\\sf \cos(\theta)=\dfrac{A}{H}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\\\phantom{ww}\bullet\;\textsf{$H$ is the hypotenuse (the side opposite the right angle).}\end{array}}[/tex]
In this case:
Substitute the values into the cosine ratio and solve for H:
[tex]\cos 42^{\circ}=\dfrac{15.2}{H}\\\\\\H=\dfrac{15.2}{\cos 42^{\circ}}\\\\\\H=20.4536174900...[/tex]
Finally, to find the perimeter, we sum the side lengths:
[tex]\textsf{Perimeter}=15.2+O+H\\\\\textsf{Perimeter}=15.2+13.686141473327...+20.4536174900...\\\\\textsf{Perimeter}=49.3397589633...\\\\\textsf{Perimeter}=49.3\; \sf m\;(1\;d.p.)[/tex]
Therefore, the perimeter of the given right triangle rounded to one decimal place is:
[tex]\LARGE\boxed{\boxed{49.3\; \sf m}}[/tex]
[tex]\dotfill[/tex]
Additional Notes
If we round each side length to 1 decimal place before adding the side lengths together, the total perimeter will be 13.7 + 20.5 + 15.2 = 49.4 m. However, rounding mid-calculation is discouraged, as this often results in an inaccurate final answer.
