Chapter 19

In Fig. 19.3, find the length of \(E F\) \(d\) Fig. \(19.3\) By Pythagoras' theorem: \(e^{2}=d^{2}+f^{2}\) Hence $$ \begin{aligned} &13^{2}=d^{2}+5^{2} \\ &169=d^{2}+25 \\ &d^{2}=169-25=144 \end{aligned} $$ Thus $$ d=\sqrt{144}=12 \mathrm{~cm} $$ i.e. $$ E F=12 \mathrm{~cm} $$

Problem 6. If \(\cos X=\frac{9}{41}\) determine the value of \(\sin X\) and \(\tan X\). Figure \(19.10\) shows a right-angled triangle \(X Y Z\). Fig. \(19.10\) Since \(\cos X=\frac{9}{41}\), then \(X Y=9\) units and \(X Z=41\) units. Using Pythagoras' theorem: \(41^{2}=9^{2}+Y Z^{2}\) from which \(Y Z=\sqrt{41^{2}-9^{2}}=40\) units. Thus \(\sin X=\frac{40}{41} \quad\) and \(\quad \tan X=\frac{40}{9}=4 \frac{4}{9}\)

Problem 7. Point \(A\) lies at co-ordinate \((2,3)\) and point \(B\) at \((8,7)\). Determine (a) the distance \(A B\), (b) the gradient of the straight line \(A B\), and (c) the angle \(A B\) makes with the horizontal. (a) Points \(A\) and \(B\) are shown in Fig. 19.11(a). In Fig. 19.11(b), the horizontal and vertical lines \(A C\) and \(B C\). are constructed. (a) (b) Fig. \(19.11\) Since \(A B C\) is a right-angled triangle, and \(A C=(8-2)=6\) and \(B C=(7-3)=4\), then by Pythagoras' theorem $$ A B^{2}=A C^{2}+B C^{2}=6^{2}+4^{2} $$ and \(\boldsymbol{A B}=\sqrt{6^{2}+4^{2}}=\sqrt{52}=7 . \mathbf{2 1 1}\), correct to 3 decimal places. (b) The gradient of \(A B\) is given by \(\tan \theta\), i.e. \(\quad\) gradient \(=\tan \theta=\frac{B C}{A C}=\frac{4}{6}=\frac{2}{3}\) (c) The angle \(A B\) makes with the horizontal is given by \(\tan ^{-1} \frac{2}{3}=33.69^{\circ}\) 7\. Point \(P\) lies at co-ordinate \((-3,1)\) and point \(Q\) at \((5,-4)\). Determine (a) the distance \(P Q\), (b) the gradient of the straight line \(P Q\) and, (c) the angle \(P Q\) makes with the horizontal.

Sketch a right-angled triangle \(A B C\) such that \(B=90^{\circ}, A B=5 \mathrm{~cm}\) and \(B C=12 \mathrm{~cm}\). Determine the length of \(A C\) and hence evaluate \(\sin A, \cos C\) and \(\tan A\) Triangle \(A B C\) is shown in Fig. 19.14. By Pythagoras' theorem, \(A C=\sqrt{5^{2}+12^{2}}=13\) By definition: \(\sin A=\frac{\text { opposite side }}{\text { hypotenuse }}=\frac{\mathbf{1 2}}{\mathbf{1 3}}\) or \(\mathbf{0 . 9 2 3 1}\) \(\cos C=\frac{\text { adjacent side }}{\text { hypotenuse }}=\frac{12}{13}\) or \(0.9231\) \(\tan A=\frac{\text { opposite side }}{\text { adjacent side }}=\frac{\mathbf{1 2}}{\mathbf{5}} \quad\) or \(\quad \mathbf{2 . 4 0 0}\)

Problem 9. In triangle \(P Q R\) shown in Fig. 19.15, find the lengths of \(P Q\) and \(P R\). Fig. \(\mathbf{1 9 . 1 5}\) $$ \begin{aligned} \tan 38^{\circ} &=\frac{P Q}{Q R}=\frac{P Q}{7.5} \\ \text { hence } \quad P Q &=7.5 \tan 38^{\circ}=7.5(0.7813)=5.860 \mathrm{~cm} \\\ \cos 38^{\circ} &=\frac{Q R}{P R}=\frac{7.5}{P R} \\ \text { hence } \quad P R &=\frac{7.5}{\cos 38^{\circ}}=\frac{7.5}{0.7880}=\mathbf{9 . 5 1 8} \mathrm{cm} \end{aligned} $$ [Check: Using Pythagoras' theorem $$ \left.(7.5)^{2}+(5.860)^{2}=90.59=(9.518)^{2}\right] $$

An electricity pylon stands on horizontal ground. At a point \(80 \mathrm{~m}\) from the base of the pylon, the angle of elevation of the top of the pylon is \(23^{\circ}\). Calculate the height of the pylon to the nearest metre. Figure \(19.23\) shows the pylon \(A B\) and the angle of elevation of \(A\) from point \(C\) is \(23^{\circ}\). $$ \tan 23^{\circ}=\frac{A B}{B C}=\frac{A B}{80} $$ Hence height of pylon \(A B=80 \tan 23^{\circ}=80(0.4245)\) ) $$ =33.96 \mathrm{~m} $$ \(=34 \mathrm{~m}\) to the nearest metre

Problem 15. Evaluate correct to 4 decimal places: (a) sine \(11^{\circ}\) (b) sine \(121.68^{\circ}\) (c) sine \(259^{\circ} 10^{\prime}\) (a) sine \(11^{\circ}=\mathbf{0 . 1 9 0 8}\) (b) sine \(121.68^{\circ}=\mathbf{0 . 8 5 1 0}\) (c) sine \(259^{\circ} 10^{\prime}=\operatorname{sine} 259 \frac{10^{\circ}}{60}=-\mathbf{0 . 9 8 2 2}\)

Evaluate, correct to 4 decimal places: (a) cosine \(23^{\circ}\) (b) cosine \(159.32^{\circ}\) (c) cosine \(321^{\circ} 41^{\prime}\) (a) cosine \(23^{\circ}=\mathbf{0 . 9 2 0 5}\) (b) cosine \(159.32^{\circ}=-\mathbf{0 . 9 3 5 6}\) (c) \(\operatorname{cosine} 321^{\circ} 41^{\prime}=\operatorname{cosine} 321 \frac{41^{\circ}}{60}=\mathbf{0 . 7 8 4 6}\)

Evaluate, correct to 4 significant figures: (a) tangent \(276^{\circ}\) (b) tangent \(131.29^{\circ}\) (c) tangent \(76^{\circ} 58^{\prime}\) (a) tangent \(276^{\circ}=\mathbf{- 9 . 5 1 4}\) (b) tangent \(131.29^{\circ}=-\mathbf{1 . 1 3 9}\) (c) \(\operatorname{tangent} 76^{\circ} 58^{\prime}=\tan 76 \frac{58^{\circ}}{60}=\mathbf{4 . 3 2 0}\)

Evaluate, correct to 4 significant figures: (a) \(\sin 1.481\) (b) \(\cos (3 \pi / 5)\) (c) \(\tan 2.93\) (a) \(\sin 1.481\) means the sine of \(1.481\) radians. Hence a calculator needs to be on the radian function. Hence \(\sin 1.481=\mathbf{0} . \mathbf{9 9 6 0}\) (b) \(\cos (3 \pi / 5)=\cos 1.884955 \ldots=-0.3090\) (c) \(\tan 2.93=-0.2148\)

Determine the acute angles: (a) \(\sin ^{-1} 0.7321\) (b) \(\cos ^{-1} 0.4174\) (c) \(\tan ^{-1} 1.4695\) (a) \(\sin ^{-1} \theta\) is an abbreviation for 'the angle whose sine is equal to \(\theta\) '. \(0.7321\) is entered into a calculator and then the inverse sine (or \(\left.\sin ^{-1}\right)\) key is pressed. Hence \(\sin ^{-1} 0.7321=47.06273 \ldots\) Subtracting 47 leaves \(0.06273 \ldots .{ }^{\circ}\) and multiplying this by 60 gives \(4^{\prime}\) to the nearest minute. Hence \(\sin ^{-1} 0.7321=47.06^{\circ}\) or \(47^{\circ} 4^{\prime}\) Alternatively, in radians, \(\sin ^{-1} 0.7321=0.821\) radians. (b) \(\cos ^{-1} 0.4174=65.33^{\circ}\) or \(65^{\circ} 20^{\prime}\) or \(\mathbf{1 . 1 4 0}\) radians. (c) \(\tan ^{-1} 1.4695=55.76^{\circ}\) or \(55^{\circ} 46^{\prime}\) or \(0.973\) radians.

Evaluate the following expression, correct to 4 significant figures: $$ \frac{4.2 \tan 49^{\circ} 26^{\prime}-3.7 \sin 66^{\circ} \mathrm{l}^{\prime}}{7.1 \cos 29^{\circ} 34^{\prime}} $$ By calculator: $$ \tan 49^{\circ} 26^{\prime}=\tan \left(49 \frac{26}{60}\right)^{\circ}=1.1681 $$ \(\sin 66^{\circ} 1^{\prime}=0.9137\) and \(\cos 29^{\circ} 34^{\prime}=0.8698\) $$ \text { Hence }=\frac{4.2 \tan 49^{\circ} 26^{\prime}-3.7 \sin 66^{\circ} 1^{\prime}}{7.1 \cos 29^{\circ} 34^{\prime}} \\ { } &{=\frac{(4.2 \times 1.1681)-(3.7 \times 0.9137)}{(7.1 \times 0.8698)}} \\\ { } &{=\frac{4.9060-3.3807}{6.1756}=\frac{1.5253}{6.1756}} \end{array} $$ \(=0.2470=0.247\), correct to 3 significant figures.

Evaluate correct to 4 decimal places: \(\begin{array}{ll}\text { (a) } \sin \left(-112^{\circ}\right) & \text { (b) cosine }\left(-93^{\circ} 16^{\prime}\right)\end{array}\) (c) tangent \(\left(-217.29^{\circ}\right)\) (a) Positive angles are considered by convention to be anticlockwise and negative angles as clockwise. From Fig. \(19.26,-112^{\circ}\) is actually the same as \(+248^{\circ}\) (i.e. \(\left.360^{\circ}-112^{\circ}\right)\) Hence, by calculator, \(\sin \left(-112^{\circ}\right)=\sin 248^{\circ}=-\mathbf{0 . 9 2 7 2}\)