Question

on edge AB of a wall of a house a projection AP which is perpendicular to the wall iserected and its edga is of length 27 cm A point searchlight is fixed on the corner point p of project which a rays of light PQ such that AQ=9root3 cm if ApQ =thata then find the value of 1thata and 2tan thata+ sec thata​

Answer 1

Given : AB of a wall of a house  

searchlight is fixed on the corner point p which a rays of light PQ

AQ = 9√3

AP = 27 cm

∠APQ   = θ

To Find :   θ

2tan θ + sec θ

Solution:

tan θ =  AQ/AP

=> tan θ =  9√3/27

=> tan θ =   √3/3

=> tan θ =   1/√3

=> θ =  30°

Cos30° = 1/2  => Sec30° = 2

2tan θ + sec θ  = 2(1/√3)  +  2

=  (2/√3)    + 2

= 2√3/3  + 2

= (6 + 2√3)/3

2tan θ + sec θ  = (6 + 2√3)/3

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Answer 2

Answer:

above answer is correct