Given that the distance and angle of elevation of the tree from P are 50 m. and 32°, and Q is 100 m. from P, we have;
a. The height of the tree is approximately 31.2 m.
b. i. The distance of Q from the base of the tree is 50•√5 m.
ii. The angle of elevation of the top of the tree from Q is approximately 15.6°
iii. The bearing of the tree from Q is 296.565°
How can the distances and bearing of the tree from Q be found?
a. tan(32°) = Height/50
Height = 50 × tan(32°) = 31.243 m.
- Height of the tree ≈ 31.2 m.
b. i. The distance, d, of Q from the base of the tree is given by Pythagorean theorem as follows;
d = √(100² + 50²) = 50•√5
- The distance of Q from the base of the tree, d = 50•√5 m.
ii. The angle of elevation is given by trigonometric ratios as follows;
[tex]tan (\theta) = \frac{31.243}{50 \cdot \sqrt{5} } [/tex]
Which gives;
[tex]\theta= arctan \left( \frac{31.243}{50 \cdot \sqrt{5} } \right) \approx 15.6 ^{ \circ} [/tex]
- The angle of elevation of the top of the tree from Q ≈ 15.6°
iii. The angle formed at Q by the lines from Q to the point P and Q to the tree is found as follows;
Angle = arctan(50/100) ≈ 26.565°
The angle between the direction north of Q and a line from Q to the tree is therefore;
90° - 26.565° = 63.435°
The bearing, which is the angle described clockwise from the direction north of Q and the direction from Q to the tree, is therefore;
Bearing = 360° - 63.435° = 296.565°
- The bearing of the tree from Q = 296.565°
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