Determining the component form of a vector.
Determining the magnitude and direction of a vector.
Determining the magnitude and direction of a vector.
Use the Law of Sines to solve oblique triangles
Use the Law of Cosines to solve oblique triangles.
Using the Heron’s Formula to determine the area of an oblique triangle.
Using the sum and difference formulas to determine exact values of trigonometric expressions and verify identities.
100
Given a vector with initial point (−2, 8) and terminal point (4, 3), find an equivalent vector whose initial point is (0, 0). Write the vector in the component form 
a, b
.
6, -5
Explanation:
A vector in standard position originates from the origin. For this, we will use the component form and subtract x2-x1, y2-y1,
4-(-2), 3-8 equals 6, -5
100
What is the Law of Sines?
(Sin A / a) =(Sin B / b)
100
What is the Law of Cosines?
a2 = b2+c2-2bccos A
100
The square root of s(s-a)(s-b)(s-c)
What is Heron's Formula for the area of a triangle?
100
sinAcosB +cosAsinB is equal to
Sin(A+B)
200
Given initial point P1 = (-3,5) and terminal point P2 = (6, 4), write the vector v in terms of i and j.
v = 9i-j
Explanation:
To write a vector in terms of i and j, we use the formula ai + bj =v.
6-(-3), 4-5
200
Given two sides and one opposite angle, to solve a triangle you will..
Use the Law of Sines and check for a second possible triangle.
200
Assume α is opposite side a, β is opposite side b, and γ is opposite side c. If possible, solve the triangle for the unknown side, c. Round to the nearest tenth.
γ = 52.3°, a = 9.79, b = 3.33
c is approximately 8.2.
Use the formula to solve.
c2= b2 +a2-2bacos C
c2 = (9.79)2 + (3.33)2 - 2(9.79)(3.33)cos(52.3)
c=8.2
200
How do you find the variable s in Heron's Formula?
(a + b + c) / 2
a, b, and c are all sides of a triangle.
200
cos(A-B) is equal to
cosAcosB + sinAsinB
300
Find the magnitude of the vector, 0 ≤ θ < 2π.
6, -4
the square root of {52}
Explanation:
To find the magnitude of a vector, use the formula:
v = the square root of {a2 + b2}
6 x 6 = 36
-4 x -4 = 16
36 +16 = 52
52 is not a perfect square, so we leave it as the square root of 52.
300
Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Round your answer to the nearest tenth. Find c, if possible.
α = 40°, γ =120°, a = 12
c =16.2
To solve an AAS triangle, we need to use the Law of Sines.
Sin(40) / 12 = Sin(120) / c
12sin(120) = cSin(40)
c=16.2
300
Use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.
Find angle A.
a= 9 b= 7.2 c= 6.
angle A = 85.46
Use the Formula
a2 = b2+c2-2bccos A
92=7.22+62-2(7.2)(6)cosA
81=87.84-86.4cosA
-6.84=-86.4cosA
arccosine(.07917)= 85.46
300
Find the area of a triangle to the nearest hundredth.
a = 3
b = 9
c =11
26.98 units2
(3+9+11)/2=13 =s
(13)(13-3)(13-9)(13-11)=728
square root of 728= 26.98 units2
300
What is the exact value of
cos(19π/12)
(square root 6 - square root 2) / 4
cos(19π/12) = (22π/12 - 3π/12) = ( 11π/6 - π/4)
(square root 3 / 2)(square root2/2) + (-1/2)(square root 2/2)
400
Find the direction of the vector, 0 ≤ θ < 2π.
5, 8
arctan(8/5)
Explanation:
To find the direction of a vector, use the formula arctan(b/a)
400
Find angle B when C= 82°, b =11, c = 7.
IMPOSSIBLE
In an ambiguous case, a triangle with SSA, you may have to check for 2 different possible triangles. In this case, the triangle is not possible because when the sin(theta) is greater than 1, the triangle is deemed impossible. When the sin(theta) is less than 1, the triangle is possible and we should find an acute and an obtuse angle.
400
Given two sides and one included angle, to solve triangle you will use the Law of Sines and check for a second triangle.
True or False?
False.
When given two sides and an included angle, you should use the Law of Cosine. Use the Law of Sines when given two sides and an opposite angle.
400
Solve the area of the triangle using the sides given. Round to the nearest hundredth.
a = 5
b =2.5
c=9
IMPOSSIBLE
(2.5+5+9)/2= 8.25
(8.25)(8.25-5)(8.25-2.5)(8.25-9)=-275.73
We cannot take the square root of a negative number, so the triangle does not exist. Area of a triangle cannot be negative.
400
Rewrite in terms of sin(x) and cos(x).
cos ( 45 - x)
((square root 2)/2)cos(x)+ ((square root 2)/2)sin(x)
solve by plugging in variables to the formula for Cos(A-B)
500
Using the given magnitude and direction in standard position, write the vector in component form. Round to the nearest tenth.
|v| = 16, θ = 45°
-2.6, 14.8
Explanation:
To write a vector in component form using only magnitude and direction, use the rules:
a= (the absolute value of V)cosine(angle theta)
b= (the absolute value of V)sine(angle theta)
15cos(100)= -2.6
15sin(100)= 14.8
500
Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve for a, if possible. Round your answer to the nearest tenth.
α = 46°, β = 39°, b= 8
a = 9.1
Explanation:
Use the law of sines.
Sin(46) / a = Sin(39) / 8
8Sin(46) = a Sin(39)
a = 9.1
500
Solve the angles of the triangle. Round to the nearest hundredth.
a= 4
b=8
c=9
A= 26.38 degrees
B= 77.16 degrees
C= 76 degrees
42=82+92-2(8)(9)cosA
16=145-144cosA
-129/-144=cosA
arccosine(.89583)= 26.28 =A
82=42+92-2(4)(9)cosB
81=97-72cosB
-16/-72=cosB
Arccosine(.2222)=77.16=B
180-26.38-77.16= 76 = C
500
Solve the area of the triangle with the information given to the nearest hundredth.
a = 12
b= 12
C= 24.1
29.34 units2
Use the Law of Cosine to solve fro the missing side. Use the Heron's formulas to find the area.
c2=122+122-2(12)(12)cos(24.1)
c=5
(12+12+5) / 2 =14.5
14.5(14.5-12)(14.5-12)(14.5-5)= 860.9375
square root (860.9375) = 29.34
500
Find the exact value algebraically.
tan (255)
-2 - (square root of 3)
Use the formula for tangent
Tan(A+B)= (tanA +tanB)/ (1- tanAtanB)
tan (255)= Tan(225+30)
(1+ square root of 3)/(1-(1)(square root 3)