103 Trigonometry Problems

46 103 Trigonometry Problems
−−→
OM = √ 3 [ 13, a−b ] [
2 . Hence, a+b
2 , 24] = √ 3 [ 13, a−b ]
2 ; that is,
a + b
2
= 13 √ 3 and
a − b
2
= 8 √ 3.
Adding the last two equations gives a = 21 √ 3, and subtracting the second equation
from the first equation gives b = 5 √ 3. Consequently, ab = 315.
Second Solution: Let ̸ α denote the angle formed by ray OA and the positive
direction of the x axis, and set x =|OA| =|OB| =|AB|. Then sin α = 11
x
and
cos α =
x a . Note that ray OB forms an angle whose measure is α + 60◦ from the
positive x axis. Then by the addition and subtraction formulas, wehave
37
x = sin(α + 60◦ ) = sin α cos 60 ◦ + cos α sin 60 ◦ = 11
2x + a√ 3
2x ,
b
x = cos(α + 60◦ ) = cos α cos 60 ◦ − sin α sin 60 ◦ = a
2x − 11√ 3
2x .
Solving the first equation for a gives a = 21 √ 3. We then solve the second equation
for b to obtain b = 5 √ 3. Hence ab = 315.
The Dot Product and the Vector Form of the Law of
Cosines
In this section we introduce some basic knowledge of vector operations. Let u =
[a,b] and v =[m, n] be two vectors. Define their dot product u · v = am + bn. It
is easy to check that
(i) v · v = m 2 + n 2 =|v| 2 ; that is, the dot product of a vector with itself is
the square of the magnitude of v, and v · v ≥ 0 with equality if and only if
v =[0, 0];
(ii) u · v = v · u;
(iii) u · (v + w) = u · v + u · w, where w is a vector;
(iv) (cu) · v = c(u · v), where c is a scalar.
If vectors u and v are placed tail to tail at the origin O, let A and B be the heads of u
and v, respectively. Then −→ AB = v − u. Let θ denote the angle formed by lines OA
and OB. Then ̸ AOB = θ. Applying the law of cosines to triangle AOB yields
|v − u| 2 = AB 2 = OA 2 + OB 2 − 2OA · OB cos θ
=|u| 2 +|v| 2 − 2|u||v| cos θ.