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Ivanshal [37]
3 years ago
5

Company A is offering 2 1/2 pounds of chocolate for $32.50, while Company B is offering 2 3/4 of the same chocolate for $35.50

Mathematics
2 answers:
yuradex [85]3 years ago
6 0

Hi! :D The answer I got is Company B is cheaper. The way I got the answer is I did 32.50/2.5=$13(Company A) and for Company B I did 35.50/2.75= 12.90 approximately. So obviously 12.90 is much cheaper than 13 even though it is 10 cents difference.

I hoped I helped! ;D

Ivan3 years ago
3 0
Company B offers the lowest price.

Company A:
32.50/2.5 =
13

Company B:
35.50/2.75 =
12.90

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zlopas [31]

Answer:

Solution to Problem 1:

Use the fact that the sum of all three angles of a triangle is equal to 180 o to write an equation in C.

A + B + C = 180 o

Solve for C.

C = 180 o - (A + B) = 43 o

Use sine law to write an equation in b.

a / sin(A) = b / sin(B)

Solve for b.

b = a sin (B) / sin(A) = (approximately) 5.4 cm

Use the sine law to write an equation in c.

a / sin(A) = c / sin(C)

Solve for c.

c = a sin (C) / sin(A) = (approximately) 7.1 cm

Problem 2

The angle of elevation to the top C of a building from two points A and B on level ground are 50 degrees and 60 degrees respectively. The distance between points A and B is 30 meters. Points A, B and C are in the same vertical plane. Find the height h of the building(round your answer to the nearest unit).

diagram problem 2

Solution to Problem 2:

We consider triangle ABC. Angle B internal to triangle ABC is equal to

B = 180 o - 60 o = 120 o

In the same triangle, angle C is given by.

C = 180 o - (50 o + 120 o) = 10 o

Use sine law to find d.

d / sin(50) = 30 / sin(10)

Solve for d.

d = 30 *sin(50) / sin(10)

We now consider the right triangle.

sin (60) = h / d

Solve for h.

h = d * sin(60)

Substitute d by the expression found above.

h = 30 *sin(50) * sin(60) / sin(10)

Use calculator to approximate h.

h = (approximately) 115 meters.

Problem 3

A triangle ABC has side a = 12 cm, side b = 19 cm and angle A = 80 o (angle A is opposite side a). Find side c and angles B and C if possible.(round answers to 1 decimal place).

Solution to Problem 3:

Use sine law to write an equation in sin(B).

a / sin(A) = b / sin(B)

Solve for sin(B).

sin (B) = (b / a) sin(A) = (19/12) sin(80) = (approximately) 1.6

No real angle B satisfies the equation

sin (B) = 1.6

The given problem has no solution.

Problem 4

A triangle ABC has side a = 14 cm, side b = 19 cm and angle A = 32 o (angle A is opposite side a). Find side c and angles B and C if possible.(round answers to 1 decimal place).

Solution to Problem 4

Use sine law to write an equation in sin(B).

a / sin(A) = b / sin(B)

Solve for sin(B).

sin (B) = (b / a) sin(A) = (19/14) sin(32) = (approximately) 0.7192

Two angles satisfy the equation sin (B) = 0.7192 and the given problem has two solutions

B1 = 46.0 o and B2 = 134 o

Solution 1: Find angle C1 corresponding to B1

C1 = 180 - B1 - A = 102 o

Solution 1: Find side c1 corresponding to C1

c1 / sin(C1) = a / sin(A)

c1 = 14 sin(102) / sin(32) = (approximately) 25.8 cm

Solution 2: Find angle C2 corresponding to B2

C2 = 180 - B2 - A = 14 o

Solution 2: Find side c2 corresponding to C2

c2 / sin(C2) = a / sin(A)

c1 = 14 sin(14) / sin(32) = (approximately) 6.4 cm

Exercises

1. A triangle ABC has angle A = 104 o, angle C = 33 o and side c = 9 m. Solve the triangle ABC by finding angle B and sides a and b.(round answers to 1 decimal place).

2. Redo problem 2 with the distance between points A and B equal to 50 meters.

Solutions to Above Exercises

1. B = 43 o, a = 16.0 m , b = 11.3 m

2. 191 meters.

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