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The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
Download (direct link): theatozofmath2002.pdf
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Substituting b = 8.4, c = 7.3, and A = 84°
Multiplying terms, and rounding to 2 dp
Adding and subtracting the terms Taking the square root
COSINE RULE 125
c =4.1. We are given the lengths of three sides, therefore we use the cosine rule to find angle C. Write
a
+ b2
cos C =---------------- Cosine rale for finding angle C
2 ab
32 _|_ 22 — 4.12
cos C = — Substituting a = 3, b = 2, and c = 4.1
2x3x2
cos C = —0.3175 Squaring terms and simplifying
A
(d)
The negative sign indicates that angle C is obtuse:
C = cos-1 (—0.3175) Use inv cos on the calculator
= 108.5° (to 1 dp)
Angle PRQ = 108.5° Replacing Cby PRQ
A trigonometric formula for the area of a triangle is introduced here. When two sides and the angle between the two sides of a triangle are known, the area of the triangle can be found using a trigonometric formula. Using the triangle notation where the three angles are A, B, and C and the lengths of the sides are a, b, and c, we have three versions of this area formula for the triangle, and we choose the appropriate one according to the information given about the triangle.
The three versions of the formula for the area of any triangle are
Area = |be sin A, Area = |ab sin C, Area = |ac sin B
Example 3. John makes gourmet cheeses and he has designed eight triangles of cheese to fit into an octagonal box. The longest diagonal of the box is 10 cm. Find the area of the lid of the box.
Solution. The lid is made up of eight congruent triangles (see figure d) and one of them is triangle ABC. The area of the lid is eight times the area of triangle ABC.
126 COST PRICE
(e)
In triangle ABC, BC
5 cm
Half the longest diagonal of the octagon.
a = 5 cm b = 5 cm Angle ACB = 45°
C =45°
Using the triangle notation. a = b
360° in a full turn 8 angles = 45° Angle ACB is angle C
Choosing the appropriate formula = | x 5 x 5 x sin45° Substituting a = 5, b = 5, C = 45°
= 8.839 (to 3 dp) Using the calculator
Area of lid = 70.7 cm2 8 sectors x 8.839 = 70.7 to 1 dp
Area of triangle ABC = \ab sin C
References: Cosine, Obtuse Angle, Octagon, Sine, Sine Rule, Tangent, Trigonometry.
COST PRICE
Amanda runs a sports shop. When she buys a pair of running shoes from a manufacturer to sell in the shop the price she pays for them is called the cost price (CP). The price at which she sells them to a customer is called the selling price (SP). The difference between these two prices is called her profit or markup. This can be written as a formula:
Profit = SP - CP The formula for percentage profit is
profit
% profit = :— x 100
cost price
Example 1. Amanda buys in a pair of golf shoes for $189 and sells them to a customer for $289.50. Find her percentage profit on the golf shoes.
COST PRICE 127
Solution. Write
Profit = $289.50 — $189.00 Using the formula Profit = SP — CP Profit = $100.50
100.50
% profit = x 100
1 189
Using the formula % profit
cost price
% profit = 53.2% (to 1 dp) Using the calculator
The problem may be restated, as in the following example, to find the selling price.
Example 2. Amanda buys in a netball from the supplier for $38. For how much should she sell it to make a profit of 64%?
Solution. We can regard the cost price as 100% and the selling price as 100 + 64 = 164%. Write
Selling price = 164% x cost price
The selling price is $63.32.
The problem may be restated again to find the cost price, as in the next example.
Example 3. Amanda puts a markup of 60% on a baseball bat, which she then sells for $152. What is the cost price?
Solution. The cost price is 100% and the selling price is (100 + 60) = 160%. Write 100
Cost price = x selling price
164% = —— and CP = $38
SP = 63.32
Using the calculator
CP = 95
CP = 0.625 x 152
SP = $152 and = 0.625
160
Using the calculator.
The cost price of the baseball bat is $95.
Reference: Percentage.
128 CRITICAL POINT
COUNTEREXAMPLE
This is a particular example which, if it can be found, is used to disprove a formula or theory which is generally believed to be true. A general theory cannot be true if there is at least one example where the theory fails to be true.
Example. It has long been the dream of mathematicians to get a formula for prime numbers. One such formula put forward was P = n2 —n + 41, where n is any counting number and P is a prime number. Find a counterexample that disproves this formula.
Solution. To find a counter example we can substitute any of the counting numbers n = 1, 2, 3,... into the formula until a value for P turns up which is not a prime number:
n = 1 P = l2 - 1 + 41
P =41 Which is a prime number
n =7 P =72 -7 +41
P = 83 Which is a prime number
n = 13 P = 132 - 13 + 41
P = 197 Which is a prime number
n = 41 P =412-41 +41
P = 1681 Which is not a prime number
1681 is not a prime number, because it has three factors, which are 1, 41, and 1681. A counterexample is when n = 41, which disproves the formula.
References: Formula, Prime Numbers, Substitution.
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