Engineering and Construction F3HV 11 Maths Craft 1 RIGHT ANGLED TRIANGLES : PYTHAGORAS' THEOREM

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1 RIGHT NGLED TRINGLES : PYTHGORS' THEOREM ver important triangle, in terms of practical use, is the right-angled triangle. In the "real" world, the right-angled triangle is used etensivel. It is a shape that has so man uses and applications that the relationships concerning its sides and angles are, effectivel, compulsor learning for an course of stud that involves mathematics that ma involve practical applications. Pthagoras' Theorem Pthagoras was an ancient Greek philosopher and mathematician. He is credited with formulating the relationship between the lengths of the sides of the right-angled triangle, but the relationship had been used b ancient Greek and Egptian builders long before Pthagoras made his formal, mathematical statement of the relationship. The formal statement of Pthagoras' Theorem states The square on the hpotenuse of a right-angled triangle equals the sum of the squares on the other two sides B c hpotenuse a b fig 1 C From this statement of the relationship between the three sides of a right-angled triangle can be written a formula which, in turn, can be used to solve problems. The eact form of the equation depends entirel upon the triangle being used and the identification of the sides of the triangle in question. There is no particular formula that is the statement of Pthagoras' Theorem; a particular formula is the result of appling Pthagoras' Theorem to a particular triangle. The hpotenuse of a right-angled triangle is the side that is opposite the right angle; in the triangle shown above in fig 1, the hpotenuse is side a, or BC. 3

2 The formula that appling Pthagoras' Theorem to the triangle in fig 1 produces is a 2 b 2 + c 2 or BC 2 B 2 + C 2 Once the particular formula for the particular triangle has been established, then the unknown length of the third side of the right-angled triangle can be found from the known lengths of the other two sides. e.g. Given that B 5 cm, and C 8 cm, use Pthagoras' Theorem on the triangle in fig 1 to find BC. From fig 1, BC 2 B 2 + C 2 (5) 2 + (8) BC cm e.g. Given that B 12 mm, and BC 26 mm, use Pthagoras' Theorem on the triangle in fig 1 to find C From fig 1, BC 2 B 2 + C 2 i.e. C 2 BC 2 B 2 (26) 2 (12) C mm Pthagoras' Theorem can be used to check whether a triangle is right-angled, or not. Given the lengths of the sides of a triangle, square these lengths. dd the two smaller values. If this sum equals the third value, the triangle is right-angled, with the right angle opposite the third side. If the sum does not equal the third value, the triangle is not right-angled. e.g. Is the triangle with sides of lengths 12, 16, 21 a right-angled triangle? First, ; Second ; Third ; the triangle is not right-angled. (However, triangle 12, 16, 20 is right-angled since ) There are several well-known and well-used right-angled triangles that have integer values for the lengths of their sides. The first, and possibl the most famous, of these triangles is the triangle. Two other common ones are the triangle and the triangle. These three triangles actuall form bases for whole families of triangles since an triangle with sides that are multiples of one of these triangles is also a right-angled triangle. Thus, is a right-angled triangle since it is Similarl, 3½-12-12½ is a right-angled triangle since this is ½ 7-½ 24-½ 25. There are man more families of right-angled triangles formed b different base triangles with integer length sides. 4

3 Eercise 5 Find the unknown sides in the following right-angled triangles. Where eact calculations are not possible, give answers to three decimal places. (1) (2) 10cm (3) 4cm (4) (5) 12 cm 6 cm 5 cm 7 cm 9 cm 13cm (6) 15 cm 8cm (7) 24 cm (8) 2 cm 14.4cm (9) 4cm 50 cm 40cm (10) 6 cm 6cm 13cm 4cm 5

4 RIGHT NGLED TRINGLES : TRIGONOMETRIC RTIOS n angle between 0 and 90 can be part of an number of triangles, particularl right-angled triangles. However, all the right-angled triangles of which a particular angle can be a part can be placed so that smaller ones fit inside larger ones, as shown in fig 1. C N Q Y In the four right-angled triangles shown in fig 1, triangles BC, MN, PQ, and XY, the angle is the same size. The things that are the same for each triangle are the ratios of an two, corresponding sides in an triangle, e.g. the ratios B, M, P, and X are all equal. C N Q Y B M fig 1 P X There are si possible ratios that can be formed using the three sides of a rightangled triangle, each particular value of such ratios corresponding to onl one angle in a right-angled triangle. The triangle in fig 2 shows how the sides of a right-angled triangle are named. The hpotenuse is fied; it is alwas the side opposite the right angle. The other two sides are named with reference to the particular angle of concern within the triangle, in this case angle ; the opposite side is the side opposite the angle; the adjacent is the side that is net to the angle but is not the hpotenuse. Each of the si ratios has a particular name that signifies which sides are used, and how the are used, to form the ratio. These ratios are, together, the Trigonometric Ratios (or, simpl, the trig ratios) of a particular angle. For the angle in fig 2, the si trig ratios are opposite sine of, shortened to sin, hpotenuse cosine of adjacent, shortened to cos, hpotenuse tangent of opposite, shortened to tan, adjacent secant of hpotenuse adjacent cosecant of cosec,, shortened to sec, hpotenuse opposite, shortened to cotangent of adjacent, shortened to cot. opposite Of these si ratios, the three in common use are the sine, cosine and tangent ratios; the other three ratios are remembered and evaluated as being 'one over', i.e. the reciprocal of, one of the others, specificall, sec 1, cosec 1, cos sin hpotenuse cot 1. tan adjacent fig 2 opposite 6

5 Each of these ratios is a number that is related to a specific angle between 0 and 90 ; each angle between 0 and 90 has si numbers related to it, one for each of the si ratios. To find the sine, cosine, and tangent values for a particular angle then simpl use the sin, cos, and tan buttons on a calculator. Eactl how these buttons work depends on the particular machine; with some machines it is necessar to input the angle before pressing the appropriate button, for other machines, press the button first, then the angle, then ''. What is important no matter which machine is used is that the calculator is set for angles measured in degrees. The units for measuring angles is indicated on most machines b the displa showing 'DEG' or 'D'; if 'RD' (or 'R') or 'GR' (or 'G' or 'GRD') is shown, then the displa needs to be changed. To do this either read the calculator's instruction manual, or ask. If 'RD' or 'GR' (or the alternatives) are shown in the displa, the calculator will not return the correct values for the trig ratios of the angles. Eercise 6 Give the values of sin, cos, and tan for each of the following angles. (1) 30 (2) 45 (3) 60 (4) 0 (5) 90 (6) 57 (7) 14.5 (8) 6.3 (9) 77.1 (10) (If the results obtained do not agree with answers given, check that the calculator displa is set for degrees and then check that the button pressing sequence is the right one for the calculator being used.) To find the values of the other three trig ratios, again use the calculator in conjunction with the sin, cos, and tan buttons. The long wa to do this is to set up the reciprocals as 'sums'; for eample, to calculate the sec value of an angle set up 1 cos(angle). The quick wa to find the reciprocal ratios is to find cos (or sin or tan, as appropriate) and then use the 1/ (or 1 ) button on the calculator. Thus, sec25 1 cos , or sec25 co s25 [1/ ] Eercise 7 Give the values of sec, cosec, and cot for each of the following angles. (1) 30 (2) 45 (3) 60 (4) 0 (5) 90 (6) 57 (7) 14.5 (8) 6.3 (9) 77.1 (10)

6 Using trigonometric ratios to find the sides of right-angled triangles Each of the trig ratios connects one angle of a right-angled triangle to two of the sides of that triangle. If an angle and one side of the triangle are known, the lengths of the other two sides can be calculated using an appropriate trig ratio. The results of such calculations can be checked b making sure that Pthagoras' Theorem holds for the lengths calculated. The main difficulties involved with using trig ratios is remembering which ratio is which and then deciding which ratio is the appropriate one to use. useful aidememoir for making these decisions is the artificial word SOH-CH-TO (it is pronounced rather like the volcano that erupted man ears ago, Krakatoa). SOH gives Sine is Opposite over Hpotenuse; CH gives Cosine is djacent over Hpotenuse; TO gives Tangent is Opposite over djacent. To use this 'word' to decide which ratio to use, (1) mark off the S, C, and T as there must be an angle as part of the information; (2) with respect to this angle, mark off the side of known length; (3) mark off the side to have its length calculated. There should be onl one of the three sllables with all three letters marked off; this is the ratio to be used. Having decided which ratio to use, set up the equation given b the definition of the trig ratio, transpose it to make the unknown length the subject, then evaluate for the length required. e.g. In triangle BC, angle 71, angle B 90, and the hpotenuse, C, 25 mm. Calculate the lengths of the other two sides. From the diagram, fig 3, and with respect to angle, side BC is the Opposite, while side B is the djacent. Given an angle, mark off S, C, and T SOH CH TO Given hpotenuse, mark off H SOH CH TO C Want BC, opposite side, mark off O SOH CH TO Since SOH is the onl sllable completel marked off, use the sine ratio to calculate the length of BC. 25 mm To find BC Using SOH CH TO sin 71 BC BC C o BC 25sin mm fig 3 B Given an angle, mark off S, C, and T SOH CH TO Given hpotenuse, mark off H SOH CH TO Want B, adjacent side, mark off SOH CH TO Since CH is the onl sllable completel marked off, use the cosine ratio to calculate the length of B. To find B Using SOH CH TO cos 71 B B C 25 B 25cos mm 8

7 Eercise 8 Find the unknown sides in the following right-angled triangles. Where eact calculations are not possible, give our answers to four significant figures. (1) (2) (3) (4) (5) 28.5 º 13.6 cm 8.3 cm 12cm 4.7 cm 62 º 72 º 20.7 º (6) (7) (8) (9) (10) 16.7cm 72 º 16.4 cm 40.3 º 23.4 cm 6.2 cm 22 º 21.5 º 42.2 º 7.6 cm 30 º 22.6 cm Using trigonometric ratios to find the angles of right-angled triangles The first stages of finding the angle using two sides of a right-angled triangle are ver much the same as finding the length of a side. Start with SOH CH TO, mark off the S, C and T for the angle wanted, then mark off the two sides known using the angle wanted as the reference angle. Having decided on which ratio to use, the value of the ratio can be found b dividing the two known sides. To find the angle from the trig ratio value, use the calculator. ll three trig ratio buttons can invert the trig function indicated on the button; most calculators will have 'sin 1 ', 'cos 1 ', and 'tan 1 ' written above the button as a second function. These are the currentl accepted smbols for inverse sin, cos, and tan functions, i.e. the are the functions that return the angle that corresponds with the appropriate trig ratio value. e.g. sin 1 (0.45) , i.e. sin( ) cos 1 (0.78) , i.e. cos( ) 0.78 tan 1 (1.32) , i.e. tan( ) 1.32 To operate the inverse functions on a calculator press 'SHIFT' or 'INV' or '2nd' (usuall located at the top left of the button arra on most calculators) before pressing the trig function button. If the inverse functions are not written around the trig function buttons, then tr pressing the 'INV' button before the trig function button anwa. t worst, read the calculator manual. Eercise 9 Find the angles given b (1) sin 1 (0.5) (2) cos 1 (0.5) (3) tan 1 (0.5) (4) sin 1 (0.25) (5) cos 1 (0.25) (6) tan 1 (0.25) (7) sin 1 (0.75) (8) cos 1 (0.75) (9) tan 1 (0.75) (10) sin 1 (1.25) (11) cos 1 (1.25) (12) tan 1 (1.25) (13) sin 1 ( ) (14) cos 1 4 ( ) (15) tan 1 65 ( ) To find angles in a right-angled triangle, consider the following eample

8 e.g. Triangle BC is right-angled at B. Side B is 13 mm long and side C is 18 mm long. Calculate the sizes of the angles and C. Refer to fig 4. With respect to angle, side B is the djacent, while side C is the Hpotenuse. Want an angle, mark off S, C, and T SOH CH TO Given hpotenuse, mark off H SOH CH TO Given C, adjacent side, mark off SOH CH TO Since CH is the onl sllable completel marked off, use the cosine ratio to calculate the size of angle. To find angle Using SOH CH TO cos B 13 C cos angle ( ) 18 With respect to angle C, side B is the Opposite, while side C is the Hpotenuse. Want an angle, mark off S, C, and T SOH CH TO Given hpotenuse, mark off H SOH CH TO Given B, opposite side, mark off O SOH CH TO Since SOH is the onl sllable completel marked off, use the sine ratio to calculate the length of B. To find angle C Using SOH CH TO sin C B 13 C 18 angle C 1 13 ( ) sin mm 13 mm fig 4 C B Eercise 10 Find the unknown angles in the following triangles, giving our answers to the nearest tenth of a degree. (1) 18cm 12.6cm (2) 3.2cm 2.4cm (3) 7.2cm 2.43cm (4) 9.1cm 8.2cm (5) 28.3cm 24.6cm (6) 7.6cm 4cm (7) 20.4cm 7.62cm (8) 2cm (9) 3cm 7.16cm 3.41cm (10) 73.4cm 12.2cm 10

9 Isosceles triangles From the properties of isosceles triangles, right-angled triangles can be formed and the trigonometr of right-angled triangles can be put to use. The properties of the isosceles triangle to have in mind are the facts that the line from the verte to the centre of the base is perpendicular to the base and bisects the verte angle. Thus, an isosceles triangle can be split into two, identical right-angled triangles. s long as two sides and one angle or two angles and one side, together with which are the equal sides, or angles, are known, then all the missing measurements of the triangle can be calculated, along with the height of the triangle, and its area. e.g. Fig 5 shows the isosceles triangle BC, such that B C 15 cm, and angle B 42. Calculate the size of the angle s and C, the length of the side BC, and the area of the triangle. cm The line D bisects angle, bisects base BC, and is perpendicular to BC. Since B C, angle C angle B 42. angle 180 angle B angle C In right-angled triangle DB, cos42 BD BD 15cos cm o and length BC 2 BD B sin42 D D 15sin cm 15 rea of triangle 21 BC D cm cm 15 cm ns. ngle 96, angle C 42, base BC cm, area of triangle cm 2. D fig 5 C Eercise 11 (1) Find the missing sides and angles. 50 o 8 41 o o o 90 (a) (b) (c) 60 o 2 (d) 11

10 (2) Find the areas of the triangles in question 1. (3) circle of radius 6 cm encloses a regular heagon such that each verte of the heagon lies on the circumference of the circle. What is the area of the heagon? What is the unused area of the circle? [ regular heagon is a si-sided figure with all sides of equal length.] (4) n astronomer has to swing his telescope through an angle of 27.3º in order to look from one star to another. Both stars are approimatel 175 light ears from Earth. How far are the stars from each other? (5) slice of a 12" (diameter) deep pan pizza has a maimum width of 3". What is the angle at the sharp end of the pizza? 12