103 Trigonometry Problems

1. Trigonometric Fundamentals 43
y
✻
y
✻
B
A
✏ ❆❆❑ ✟
✟ ✟✟✟✟✟✟✟✟✯
✲
O
x
✏✏✏✏✏✏✏✏
✏✶
A ′
C′
✏✶
A
❃✚ ✏ ✲
✚✚✚✚✚❃ ✏✏✏✏✏✏✏✏✏✏ ✲✚ ✚✚✚✚✚❃✲ ✲
O B
B ′ x
Figure 1.44.
Note that the diagonals of a rhombus bisect the interior angles (Figure 1.44, right).
Because vectors −−→ OA ′ =|v|u and OB −−→ ′ =|u|v have the same length, we note that
if vectors −−→ OA ′ , OB −−→ ′ , and OC −−→ ′ = −−→ OA ′ + OB −−→ ′ =|v|u +|u|v are placed tail to tail,
−−→
OC ′ bisects the angle formed by vectors −−→ OA ′ and OB −−→ ′ , which is the same as the
angle formed by vectors u and v.
A vector contains two major pieces of information: its length and its direction
(slope). Hence vectors are a very powerful tool for dealing with problems in analytic
geometry. Let’s see some examples.
Example 1.14. Alex started to wander in Wonderland at 11:00 a.m. At 12:00 p.m.,
Alex was at A = (5, 26); at 1:00 p.m., Alex was spotted at B = (−7, 6). If Alex
moves along a fixed direction at a constant rate, where was Alex at 12:35 p.m.? 11:45
a.m.? 1:30 p.m.? At what time and at what location did Alex cross Sesame Street,
the y axis?
Solution: As shown in Figure 1.45, left, let A 3 ,A 1 ,A 4 denote Alex’s positions
at 12:35 p.m., 11:45 p.m., and 1:30 p.m., respectively. It took Alex 60 minutes
to move along the vector −→ AB =[−12, −20]. Hence he was dislocated by vector
−−→
AA 3 =
60
35 −→
[ ]
AB = −7, − 35 3
at 12:35 p.m. Similarly, AA −−→
1 =−60
15 −→
AB =[3, 5]
and −−→ AA 4 = 90 −→
60AB =[−18, −30]. Let O = (0, 0) be the origin. Then we find that
−−→
OA 3 = −→ OA + AA −−→
3 = [ −2, 43 ]
3 and A3 = ( −2, 43 )
3 . Likewise, A1 = (8, 31) and
A 4 = (−13, −4).
Let A 2 = (0,b) denote the point at which Alex crosses Sesame Street. Assume
that it took t minutes after 12:00 p.m. for Alex to cross Sesame Street. Then −−→ AA 2 =
t −−→
60OA 1 and OA −−→
2 = −→ OA+ −−→ AA 2 ; that is, [0,b]=[5, 26]+
60 t [−12, −20]. It follows
that [0,b]= [ 5 −
5 t , 26 − 3] t . Solving 0 = 5 −
t
5
gives t = 25, which implies that