Answer:
16.24°
Step-by-step explanation:
Are you trying to find the angle of depression from the camera to the cashier?
Assume that the camera is 2.24 m high and the cashier is 7.69 m away.
1. Angle of elevation from cashier
tanθ = 2.24/7.69 = 0.2913
θ = arctan(0.2913) = 16.24°
2. Angle of depression from camera
∠ of depression = ∠ of elevation = 16.24°
<em>2</em><em>5</em><em>%</em><em> </em><em>of</em><em> </em><em>4</em><em>0</em>
<em>=</em><em>2</em><em>5</em><em>/</em><em>1</em><em>0</em><em>0</em><em>×</em><em>4</em><em>0</em>
=<em>0</em><em>.</em><em>2</em><em>5</em><em>×</em><em>4</em><em>0</em>
<em>=</em><em>1</em><em>0</em><em> </em><em>answer</em><em>.</em>
Answer:
the third one is hard to see i think its 77 so the answer is 54.75
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
cot A =
, tanA =
, cscA =
, secA = 
Consider the right side

= 
= 
=
× sinAcosA ( cancel sinAcosA )
= cos²A - sin²A
= cos2A
= left side ⇒ verified