Answer: The height of the pole is 175.97 ( approx )
Step-by-step explanation:
Let the height of the pole is x cm,
Also, let the angle of elevation of the sun to the pole at 3 pm is ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
Thus, by the question,
![tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7B%5Ctext%7B%20The%20height%20of%20the%20pole%7D%7D%7B%5Ctext%7B%20The%20height%20of%20the%20shadow%7D%7D)
( at 3pm the height of shadow = height of the pole)
![\implies tan\theta = 1](https://tex.z-dn.net/?f=%5Cimplies%20tan%5Ctheta%20%3D%201)
![\implies \theta = 45^{\circ}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctheta%20%3D%2045%5E%7B%5Ccirc%7D)
Again, according to the question,
When the angle of elevation is
,
The height of shadow = x + 95,
![\implies tan(45-12)^{\circ}=\frac{x}{x+95}](https://tex.z-dn.net/?f=%5Cimplies%20tan%2845-12%29%5E%7B%5Ccirc%7D%3D%5Cfrac%7Bx%7D%7Bx%2B95%7D)
![\implies tan 33^{\circ}=\frac{x}{x+95}](https://tex.z-dn.net/?f=%5Cimplies%20tan%2033%5E%7B%5Ccirc%7D%3D%5Cfrac%7Bx%7D%7Bx%2B95%7D)
![\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x](https://tex.z-dn.net/?f=%5Cimplies%20tan%2033%5E%7B%5Ccirc%7Dx%20%2B%2095%5Ctimes%20tan33%5E%7B%5Ccirc%7D%3Dx)
![\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}](https://tex.z-dn.net/?f=%5Cimplies%20tan%2033%5E%7B%5Ccirc%7Dx%20-%20x%20%3D%20-%2095%5Ctimes%20tan33%5E%7B%5Ccirc%7D)
![\implies -0.3505924068 x = - 61.6937213538](https://tex.z-dn.net/?f=%5Cimplies%20-0.3505924068%20x%20%3D%20-%2061.6937213538)
![x=175.969930202\approx 175.97\text{ cm}](https://tex.z-dn.net/?f=x%3D175.969930202%5Capprox%20175.97%5Ctext%7B%20cm%7D)
-2x+9. just distribute a -1
Answer:
8r and -4r
3s and -3s
second andd fourth answer
Omg that so nice I love this app sm
Answer:
x.
Step-by-step explanation:
h(f(x) = x.
For example the inverse of f(x) = 2x is g(x) = x/2
f((gx)) 2 (x/2)
= 2x / 2
= x.