Question

Right triangles and trigonometry help!

Answer 1

Step-by-step explanation:

$sin \: 45 \degree = \frac{39}{z} \\ \\ \therefore \: \frac{1}{ \sqrt{2} } = \frac{39}{z} \\ \\ \therefore \: z = 39 \sqrt{2} \\ let \: p \: be \: the \: perpendicular \: \\ \\\therefore \: tan \: 45 \degree = \frac{39}{p} \\ \\ \therefore \:1 = \frac{39}{p} \\ \\ \therefore \:p = 39 \\ \\ sin \: 60 \degree = \frac{39}{x} \\ \\ \therefore \: \frac{ \sqrt{3} }{2} = \frac{39}{x} \\ \\ \therefore \: x = \frac{78}{ \sqrt{3} } \\ \\ \therefore \: x = \frac{78 \sqrt{3} }{ 3 } \\ \\ \therefore \: x = 26 \sqrt{3} \\\\tan\: 60 \degree = \frac{39}{y} \\ \\ \therefore \: \sqrt{3} = \frac{39}{y} \\ \\ \therefore \: y = \frac{39}{ \sqrt{3} } \\ \\ \therefore \: y= \frac{39 \sqrt{3} }{ 3 } \\ \\ \therefore \: y = 13 \sqrt{3}$