Graph y= x^2+3x+2. identify the vertex and axis of symmetry
1 answer:
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The bearing of the tree from Q is 296.565°
<h3>How to determine the height of the tree?</h3>The figure that illustrates the bearing and the distance is added as an attachment
The given parameters are:
Base of the tree, b = 50 meters
Angle (x) = 32 degrees
Calculate the height (h) of the tree using:
tan(x) = height/base
So, we have:
tan(32°) = h/50
Make h the subject
h= 50 × tan(32°)
Evaluate
h = 31.24
Hence, the height of the tree is 31.24 meters
<h3>How to determine the distance between Q and the base of the tree?</h3>The distance (d) between Q and the base of the tree
This is calculated using the following Pythagoras theorem
d = √(100² + 50²)
Evaluate
d = 111.80
Hence, the distance between Q and the base of the tree is 111.80 meters
<h3>How to determine the angle of elevation?</h3>The angle of elevation (x) using the following tangent trigonometric ratio
tan(x) = h/d
This gives
tan(x) = 31.24/111.80
Evaluate the quotient
tan(x) = 0.2794
Take the arc tan of both sides
x = 15.61
<h3>The bearing of the tree from Q </h3>This is calculated using:
Angle of bearing = 270 + arctan(50/100)
Evaluate the arc tan
Angle of bearing = 270 + 26.565
Evaluate the sum
Angle of bearing = 296.565
Hence, the bearing of the tree from Q is 296.565 degrees
Read more about bearings at:
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