Solutions to Trigonometry Problems

This JSON response contains solutions to various trigonometry problems including finding exact solutions, solving triangles, and using bearings. The solutions are provided in a step-by-step format and are suitable for quizzes and assignments. The content includes course codes and college/university names.

Quiz 3 SHOW ALL WORK to receive credit.
1. 5 pts. Solve, finding all exact solutions in
[0, 2π). 2sin2x - sin(2x) = 0
2sin^2x-sin2x=0
2sin²(x) - sin(2x) = 0
sin(x)(2sin(2x)-1) = 0
sin(x) = 0
x = 0
2sin(2x) - 1 = 0
sin(2x) = 1/2
x = π/3
x = {0, π/3}
2. 6 pts. Solve, finding all exact solutions in
[0, 2π). tan(2x + π/3) = -1
Trig table gives
tan(2x+π/3)=-1=tan(-π/4)
Trig unit circle gives another arc that has same tan value;
x=-π/4+π=3π/4
a.(2x+π/3)=-π/4→2x=-π/4−π3=π/12→x = π/24
b. 2x+π/3=3π/4→2x=3π/4−(π/3)=5π/12
x=5π/24

3. 6 pts. Solve the triangle(s), if possible. Round each final answer to the nearest hundredth.
α = 34°; a = 100; b = 120. (Side a is opposite angle α. Side b is opposite angle β.)
A=100 b=120
 
C=
a/sin A = b/sinB = c/sinC
100/sin34 = 120/sin
sin = 120* sin34/100
sin  = 120* 0.5290826861/100
sin  = o.634899216
 = arcsin 0.634899216
 = 39.410
The third angle = 180 – (39.41+34)
= 106.590
Side c
a/sinA = c/sin C
100/sin 34 = c/sin 106.59
C = 100/sin34 * sin106.59
C = 42.012
4. Side a = 25 ft., side b = 30 ft., side c = 45 ft.
4 pts. 4A. Solve the triangle. Round each answer to the nearest hundredth. 3
pts. 4B. Find the area of the triangle.
A=25 b=30
C=45
A2 = b2 + c2 – 2bccosA

252 = 302+452 – 2*30*45*cosA
625 =900+2025 – 2700cosA
625-2925 = -2700cosA
2300 = 2700cosA
Cos A = 2300/2700
cosA= 0.85185185185
angle A = arcos 0.85185185185
angle A = 31.590
a/sinA=b/sinB
25/sin 31.59 = 30/sinB
Sin B = 30* sin31.59/25
Sin B = 0.2078348142
Angle B = arcsin 0.2078348142
Angle B = 120
Angle C = 180- (12+31.59) = 136.410
Area of the triangle
According to Herons formula
Area A = square toot of (S(s-a)(s-b)(s-c))
Where S= (A+B+C)/2
S = (25+30+45)/2 = 50
Area A = square root (50(50-25)(50-30)(50-45))
Area A = square root (125000)
Area A = 353.55 square ft
For the following problem, see page 905 of the book for the definition of bearings.
5. 6 pts. Art is on a boat 4 miles east of Bea. The bearing from Art to an erupting island volcano is
N36°E. The bearing from Bea to the same volcano is N62°E. Find the distance from Bea to the volcano,
to the nearest tenth of a mile. Show a diagram of any triangles you use in your solution, with sides and
angles labeled.

Art
4 miles
26o
erupting volcano bea
considering that Art is in the west of Bea, the triangle looks like
Art
Art 4 miles Bea
1160
260
Erupting volcano
Therefore the distance of erupting volcano from bea will be
4/Sin 26 = b/sin 116
B distance = sin 116 * 4/sin 26 = 1.24 miles
30. A car travels due west for 10 miles. Then it turns and goes 2 miles in the direction of N64ᵒE. How far
is it from the starting place?
2 miles
260
Turning point start point
4 miles
C2 = a2 + b2 – 2abcosC
C2 = 22+42 – 2*4*2*cos26
C2 = 4+16-16cos26
C2 = 9.649
C=3.106
No. 29

Solve cos2x=5sinx−2
Note that cos(2x)=1−2sin2(x)
Substituting in, we have:
1−2sin2(x)=5sinx−2
Moving everything to one side:
2sin2(x)+5sinx−3=0
We now have a quadratic which we can factor:
(2sinx−1)(sinx+3)=0
So, we have that sinx=1/2 or sinx=−3, clearly the latter is not possible since sinx∈[−1,1]
Therefore x=π/6+2πk,5π/6+2πk