I’m not good at maths thank me later
So basically what you have to do is
BD is equal to 120 because when you find the hypotenuse of triangle GEB, you just have to multiply it by 1.5.
Answer:
![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Step-by-step explanation:
Given that:

recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
![f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%5C%20%20t%20%5C%20%20cos%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%20sin%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5D)
![f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%20%5C%20%20t%20%5C%20%5Cdfrac%7B3%7D%7B2%7D%2B%20sin%20%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B1%7D%7B2%7D%5D)

![L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20L%20%28%206%20%5Csqrt%7B3%7D%20%5C%20cos%20%5C%20%28t%29%20%2B%206%20%5C%20sin%20%5C%20%28t%29%20%5D)
![L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%206%20%5Csqrt%7B3%7D%20%5C%20L%20%5Bcos%20%5C%20%28t%29%20%5D%20%2B%206%5C%20L%20%5B%20sin%20%5C%20%28t%29%20%5D)



![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Answer:
I think its 30 I'm not 100% sure tho