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Ex 11.1
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Solutions
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Question 1 - Trigonometric ratios of special angles
(a)
\begin{align} {\sin 45^\circ \over \cos 30^\circ + \sin 60^\circ} & = {{1 \over \sqrt{2}} \over {\sqrt{3} \over 2} + {\sqrt{3} \over 2}} \\ & = {{1 \over \sqrt{2}} \over {2\sqrt{3} \over 2}} \\ & = {{1 \over \sqrt{2}} \over \sqrt{3}} \\ & = {1 \over \sqrt{2}} \div \sqrt{3} \\ & = {1 \over \sqrt{2}} \times {1 \over \sqrt{3}} \\ & = {1 \over \sqrt{6}} \end{align}
(b)
\begin{align} \tan 45^\circ + \tan 30^\circ \tan 60^\circ & = (1) + \left(1 \over \sqrt{3}\right)(\sqrt{3}) \\ & = 1 + \left(1 \over \sqrt{3}\right)\left(\sqrt{3} \over 1\right) \\ & = 1 + {\sqrt{3} \over \sqrt{3}} \\ & = 1 + 1 \\ & = 2 \end{align}
(c)
\begin{align} \sin {\pi \over 3} \cos {\pi \over 6} + \cos {\pi \over 3} & = \left(\sqrt{3} \over 2\right)\left(\sqrt{3} \over 2\right) + {1 \over 2} \\ & = {3 \over 4} + {1 \over 2} \\ & = {5 \over 4} \end{align}
(d)
\begin{align} {\tan {\pi \over 3} \over \sin {\pi \over 3} \tan {\pi \over 4} + \cos {\pi \over 6}} & = { \sqrt{3} \over \left(\sqrt{3} \over 2\right)(1) + {\sqrt{3} \over 2} } \\ & = { \sqrt{3} \over {\sqrt{3} \over 2} + {\sqrt{3} \over 2}} \\ & = {\sqrt{3} \over {2\sqrt{3} \over 2} } \\ & = {\sqrt{3} \over \sqrt{3}} \\ & = 1 \end{align}
Question 2 - Trigonometric ratio of complementary angles
\begin{align} \sin \theta \cos (90^\circ - \theta) & = \sin \theta(\sin \theta) \phantom{00000000} [\cos (90^\circ -\theta) = \sin \theta ] \\ & = \left(1 \over 2\right)\left(1 \over 2\right) \\ & = {1 \over 4} \end{align}
Question 3 - Trigonometric ratio of complementary angles
\begin{align} 2\tan A + \tan (90^\circ - A) & = 2\tan A + {1 \over \tan A} \phantom{00000000} \left[\tan (90^\circ - A) = {1 \over \tan A}\right] \\ & = 2(2) + {1 \over (2)} \\ & = 4{1 \over 2} \end{align}
(a)
\begin{align} \text{Since } 180^\circ < & \phantom{.} \theta < 270^\circ, \theta \text{ is in the 3rd quadrant.} \\ \\ \theta & = 180^\circ + \alpha \\ 250^\circ & = 180^\circ + \alpha \\ \\ \alpha & = 250^\circ - 180^\circ \\ & = 70^\circ \end{align}
(b)
\begin{align} \text{Since } 360^\circ < & \phantom{,} \theta < 450^\circ, \theta \text{ is in the 1st quadrant.} \\ \\ \theta & = 360^\circ + \alpha \\ 390^\circ & = 360^\circ + \alpha \\ \\ \alpha & = 390^\circ - 360^\circ \\ & = 30^\circ \end{align}
(c)
\begin{align} \text{Since } -90^\circ < & \phantom{.} \theta < 0^\circ, \theta \text{ is in the 4th quadrant.} \\ \\ \theta & = -\alpha \\ -60^\circ & = -\alpha \\ \\ \alpha & = 60^\circ \end{align}
(d)
\begin{align} \text{Since } -180^\circ < & \phantom{.} \theta < -90^\circ, \theta \text{ is in the 3rd quadrant.} \\ \\ \theta & = -(180^\circ - \alpha) \\ -100^\circ & = -180^\circ + \alpha \\ -100^\circ + 180^\circ & = \alpha \\ 80^\circ & = \alpha \end{align}
(a)
\begin{align} \theta & = 20^\circ, 180^\circ - 20^\circ, 180^\circ + 20^\circ, 360^\circ - 20^\circ \\ & = 20^\circ, 160^\circ, 200^\circ, 340^\circ \end{align}
(b)
\begin{align} \theta & = 70^\circ, 180^\circ - 70^\circ, 180^\circ + 70^\circ, 360^\circ - 70^\circ \\ & = 70^\circ, 110^\circ, 250^\circ, 290^\circ \end{align}
(c)
\begin{align} \theta & = 35^\circ, 180^\circ - 35^\circ, 180^\circ + 35^\circ, 360^\circ - 35^\circ \\ & = 35^\circ, 145^\circ, 215^\circ, 325^\circ \end{align}
(a)
\begin{align} \theta & = {\pi \over 3}, \pi - {\pi \over 3}, \pi + {\pi \over 3}, 2\pi - {\pi \over 3} \\ & = {\pi \over 3}, {2\pi \over 3}, {4\pi \over 3}, {5\pi \over 3} \end{align}
(b)
\begin{align} \theta & = {\pi \over 5}, \pi - {\pi \over 5}, \pi + {\pi \over 5}, 2\pi - {\pi \over 5} \\ & = {\pi \over 5}, {4\pi \over 5}, {6\pi \over 5}, {9\pi \over 5} \end{align}
(c)
\begin{align} \theta & = 0.3, \pi - 0.3, \pi + 0.3, 2\pi - 0.3 \\ & = 0.3, 2.8415, 3.4415, 5.9831 \\ & \approx 0.3, 2.84, 3.44, 5.98 \end{align}
Question 7 - Real-life problem
(i)
\begin{align} R & = {2(10)^2 \over (10)} \sin 30^\circ \cos 30^\circ \\ & = (20)\left(1 \over 2\right)\left(\sqrt{3} \over 2\right) \\ & = {20\sqrt{3} \over 4} \\ & = 5\sqrt{3} \end{align}
(ii)
\begin{align} R & = {2(7)^2 \over (9.8)} \sin 45^\circ \cos 45^\circ \\ & = (10)\left(1 \over \sqrt{2}\right)\left(1 \over \sqrt{2}\right) \\ & = {10 \over 2} \\ & = 5 \end{align}
(i)
\begin{align} \cos A & = {3 \over 5} \\ {Adj \over Hyp} & = {3 \over 5} \end{align}
\begin{align} x & = \sqrt{5^2 - 3^2} \\ & = 4 \\ \\ \tan A + \tan \left({\pi \over 2} - A\right) & = \tan A + {1 \over \tan A} \phantom{000000} \left[ \tan \left({\pi \over 2} - A\right) = {1 \over \tan A} \right] \\ & = {4 \over 3} + {1 \over {4 \over 3}} \\ & = {4 \over 3} + {3 \over 4} \\ & = 2{1 \over 12} \end{align}
(ii)
\begin{align} 4\sin A + \cos \left( {\pi \over 2} - A \right) & = 4\sin A + \sin A \phantom{000000} [\cos \left({\pi \over 2} - A\right) = \sin A] \\ & = 5\sin A \\ & = 5 \left(4 \over 5\right) \\ & = 4 \end{align}
(i)
\begin{align} \sin A & = {1 \over \sqrt{3}} \\ {Opp \over Hyp} & = {1 \over \sqrt{3}} \end{align}
\begin{align} \require{cancel} x & = \sqrt{ (\sqrt{3})^2 - (1)^2 } \\ & = \sqrt{2} \\ \\ \cos A \cos 30^\circ & = \left(\sqrt{2} \over \sqrt{3}\right)\left(\sqrt{3} \over 2\right) \\ & = {\sqrt{2}\cancel{\sqrt{3}} \over 2\cancel{\sqrt{3}}} \\ & = {\sqrt{2} \over 2} \end{align}
(ii)
\begin{align} \tan A \sin 45^\circ & = \left(1 \over \sqrt{2}\right)\left(1 \over \sqrt{2}\right) \\ & = {1 \over 2} \end{align}
(iii)
\begin{align} \sin (90^\circ - A) \tan 60^\circ & = \cos A \tan 60^\circ \phantom{000000} [\sin (90^\circ - A) = \cos A] \\ & = \left(\sqrt{2} \over \sqrt{3}\right)(\sqrt{3}) \\ & = \sqrt{2} \end{align}
(i)
\begin{align} A & = 180^\circ + \alpha \\ 230^\circ & = 180^\circ + \alpha \\ \\ \alpha & = 230^\circ - 180^\circ \\ & = 50^\circ \end{align}
\begin{align} B & = 50^\circ, 180^\circ - 50^\circ, 180^\circ + 50^\circ, 360^\circ - 50^\circ \\ & = 50^\circ, 130^\circ, 230^\circ, 310^\circ \end{align}
(ii)
\begin{align} A & = 360^\circ - \alpha \\ 320^\circ & = 360^\circ - \alpha \\ 320^\circ - 360^\circ & = - \alpha \\ -40^\circ & = -\alpha \\ 40^\circ & = \alpha \end{align}
\begin{align} B & = 40^\circ, 180^\circ - 40^\circ, 180^\circ + 40^\circ, 360^\circ - 40^\circ \\ & = 40^\circ, 140^\circ, 220^\circ, 320^\circ \end{align}
(i)
\begin{align} \sin A & = k \\ {Opp \over Hyp} & = {k \over 1} \end{align}
\begin{align} x & = \sqrt{(1)^2 - (k)^2} \\ & = \sqrt{1 - k^2} \\ \\ \cos A & = {\sqrt{1 - k^2} \over 1} \\ & = \sqrt{1 - k^2} \end{align}
(ii)
\begin{align} \sin (90^\circ - A) \tan 30^\circ & = \cos A \tan 30^\circ \phantom{00000000} [\sin (90^\circ - A) = \cos A] \\ & = \left(\sqrt{1 - k^2}\right) \left(1 \over \sqrt{3}\right) \\ & = {\sqrt{1 - k^2} \over \sqrt{3} } \\ & = \sqrt{ 1 - k^2 \over 3 } \end{align}
(iii)
\begin{align} {2 \cos {\pi \over 6} \over \tan (90^\circ - A)} & = {2 \cos {\pi \over 6} \over {1 \over \tan A}} \phantom{0000000000} \left[ \tan (90^\circ - A) = {1 \over \tan A} \right] \\ & = 2 \cos {\pi \over 6} \div {1 \over \tan A} \\ & = 2 \cos {\pi \over 6} \times \tan A \\ & = 2 \cos {\pi \over 6} \tan A \\ & = 2 \left({\sqrt{3} \over 2}\right) \left( k \over \sqrt{1 - k^2} \right) \\ & = \left( \sqrt{3} \over 1 \right) \left( k \over \sqrt{1 - k^2} \right) \\ & = k {\sqrt{3} \over \sqrt{1 - k^2}} \\ & = k \sqrt{ 3 \over 1 - k^2} \end{align}
\begin{align} \cos 35^\circ & = k \\ {Adj \over Hyp} & = {k \over 1} \end{align}
\begin{align} x & = \sqrt{(1)^2 - (k)^2} \\ & = \sqrt{1 - k^2} \\ \\ \tan 55^\circ & = \tan (90^\circ - 35^\circ) \\ & = {1 \over \tan 35^\circ} \phantom{000000000} \left[ \tan (90^\circ - \theta) = {1 \over \tan \theta} \right] \\ & = {1 \over {\sqrt{1 - k^2} \over k}} \\ & = 1 \div {\sqrt{1 - k^2} \over k} \\ & = 1 \times {k \over \sqrt{1 - k^2}} \\ & = {k \over \sqrt{1 - k^2}} \\ \\ \therefore \tan 35^\circ + \tan 45^\circ + \tan 55^\circ & = {\sqrt{1 - k^2} \over k} + 1 + {k \over \sqrt{1 - k^2}} \\ & = 1 + {\sqrt{1 - k^2} \over k} + {k \over \sqrt{1 - k^2}} \\ & = 1 + {\sqrt{1 - k^2}\sqrt{1 - k^2} \over k\sqrt{1 - k^2}} + {k(k) \over k\sqrt{1 - k^2}} \\ & = 1 + {1 - k^2 \over k\sqrt{1 - k^2}} + {k^2 \over k\sqrt{1 - k^2}} \\ & = 1 + {1 - k^2 + k^2 \over k\sqrt{1 - k^2}} \\ & = 1 + {1 \over k\sqrt{1 - k^2}} \end{align}