Question

$l = \frac{\frac{(H-h)}{\tan(\beta)} + \frac{(H-h)}{\tan(\gamma)} }{\cos(\psi_1 - (90 - \beta))}$

Name: l = (((H-h))/(tan(β)) + ((H-h))/(tan(γ)))/(cos(ψ1 - (90 - β)))
Uploaded: 2023-07-20T22:25:12-08:00
Duration: 1 min 51 s
Channel: Bradley Duda
Description: l = (((H-h))/(tan(β)) + ((H-h))/(tan(γ)))/(cos(ψ1 - (90 - β)))

          $l = \frac{\frac{(H-h)}{\tan(\beta)} + \frac{(H-h)}{\tan(\gamma)} }{\cos(\psi_1 - (90 - \beta))}$

Added by Ivan M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Bradley Duda

07/20/2023

Step 1

The equation can be simplified as follows: (H - h) * tan(tan^(-1)(cos(-90°/7))) = 0 Show more…

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solve this equation as a function of h H-h H-h)I tan) tan 1= cos-90- 7

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Question

$l = \frac{\frac{(H-h)}{\tan(\beta)} + \frac{(H-h)}{\tan(\gamma)} }{\cos(\psi_1 - (90 - \beta))}$

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