Trig application problem...

[FabZeros]

I would love to know how they got that identity but it's beyond me...

 

[stapel]

I would love to know how they got that identity in part (a), but it's beyond me...

A person with an eye level of 5.5 feet is standing in front of a painting. The bottom of the painting is 3 feet above floor level, and the painting is 6 feet in height. The angle \(\displaystyle \theta\) of the person's view of the painting is called "the observation angle". The person is \(\displaystyle d\) feet from the painting.

a. Verify that \(\displaystyle \theta\, =\, \tan^{-1}\left(\dfrac{6d}{d^2\, -\, 8.75}\right)\)
b. Find the distance \(\displaystyle d\) for which \(\displaystyle \theta\, =\, \dfrac{\pi}{6}\) (Round to the nearest tenth.)

What have you tried? You drew the horizontal line-of-sight line, creating two right triangles, and... then what?

Please be complete, so we can see where you're needing help. Thank you!

 

[HallsofIvy]

You will probably need the trig identity \(\displaystyle tan(\theta_1)+ tan(\theta_2)= \frac{tan(\theta_1)+ tan(\theta_2)}{1- tan(\theta_1)tan(\theta_2)}\). Notice that 8.75= (2.5)(3.5). Do you see where 2.5 and 3.5 come from?

 

[FabZeros]

I've solved it thanks to all your perfect advice! Wow I am so glad that you guys didn't just give it all away. Had you done that, I wouldn't be feeling this good right now! Thanks you guys!!!!

p.s. Should I post my steps so that other students may see?

 

[Deleted member 4993]

I've solved it thanks to all your perfect advice! Wow I am so glad that you guys didn't just give it all away. Had you done that, I wouldn't be feeling this good right now! Thanks you guys!!!!

p.s. Should I post my steps so that other students may see?

Please do post your solution - so that other students can benifit.