Question

Can someone help me solve this?​

Answer 1

Step-by-step explanation:

$sin \: 60 \degree = \frac{27}{k} \\ \\ \therefore \: \frac{ \sqrt{3} }{2} = \frac{27}{k} \\ \\ \therefore \: k = \frac{2 }{\sqrt{3} } \times 27 \\ \\ \therefore \: k = \frac{2 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3} } \times 27 \\ \\ \therefore \: k = \frac{2 \sqrt{3} }{3 } \times 27 \\ \\ \therefore \: k = 2 \sqrt{3} \times 9 \\ \\ \therefore \: k = 18 \sqrt{3} \\ \\ \therefore \: k = 18 \times 1.732 \\ \\ \therefore \: k = 31.176 \\ \huge \red{ \boxed{ \therefore \: k \approx \: 31.18 }}\\ \\ tan \: 60 \degree = \frac{27}{j} \\ \\ \therefore \: \sqrt{3} = \frac{27}{j} \\ \\ \therefore \: j = \frac{27}{ \sqrt{3} } = \frac{27 \sqrt{3} }{3} = 9 \sqrt{3} \\ \therefore \: j =9 \times 1.732 \\ \therefore \: j =15.588 \\ \huge \orange{ \boxed{ \therefore \: j \approx 15.59}}$