Algebra Quiz 3: Trigonometric Equations, Triangle Solutions, Bearings

This document presents the solutions to an algebra quiz focusing on trigonometry. It includes problems involving solving trigonometric equations for exact solutions within a specified interval, solving triangles using the Law of Sines, the Law of Cosines, and Heron's formula for area calculation, and application problems involving bearings to determine distances. The solutions are detailed and show all the necessary steps, including diagrams for better understanding of the geometric problems. Desklib provides students access to a wide range of past papers and solved assignments.

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