Exercise (10.4) - Trigonometric Ratios of Special Angles

Trigonometric Ratios of $30^{\circ}, 45^{\circ}$ and $60^{\circ}$

$\begin{array}{|c|c|c|c|c|c|c|} \hline \theta & \sin \theta & \cos \theta & \tan \theta & \cot \theta & \sec \theta & \csc \theta \\ \hline 30^{\circ}\left(\displaystyle\frac{\pi}{6}\right) & \displaystyle\frac{1}{2} & \displaystyle\frac{\sqrt{3}}{2} & \displaystyle\frac{\sqrt{3}}{3} & \sqrt{3} & \displaystyle\frac{2 \sqrt{3}}{3} & 2 \\ \hline 45^{\circ}\left(\displaystyle\frac{\pi}{4}\right) & \displaystyle\frac{\sqrt{2}}{2} & \displaystyle\frac{\sqrt{2}}{2} & 1 & 1 & \sqrt{2} & \sqrt{2} \\ \hline 60^{\circ}\left(\displaystyle\frac{\pi}{3}\right) & \displaystyle\frac{\sqrt{3}}{2} & \displaystyle\frac{1}{2} & \sqrt{3} & \displaystyle\frac{\sqrt{3}}{3} & 2 & \displaystyle\frac{2 \sqrt{3}}{3} \\ \hline \end{array}$

Exercise (10.4)

1. Draw a right triangle and find $\angle A$.
   
   (a) $\sin A=\displaystyle\frac{1}{2}$
   
   
   
   

   $ \begin{array}{l} \quad \quad \sin A=\displaystyle\frac{1}{2}\\\\ \quad \quad \displaystyle\frac{{\text{length of opposite side}}}{{\text{ length of hypotenuse side}}}=\displaystyle\frac{1}{2}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle}\text{.}\\\\ \therefore \quad \angle A=\text{30}^{\circ} \end{array}$

   
   (b) $\cos A=\displaystyle\frac{\sqrt{3}}{2}$
   
   
   
   

   $\begin{array}{l} \quad \quad\cos A=\displaystyle\frac{{\sqrt{3}}}{2}\\\\ \quad\quad \displaystyle\frac{{\text{length of adjacent side}}}{{\text{length of}\ \text{hypotenuse side}}}=\displaystyle\frac{{\sqrt{3}}}{2}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle}\text{.}\\\\ \therefore \quad \angle A=30^{\circ} \end{array}$

   
   (c) $\tan A=\sqrt{3}$
   
   
   
   

   $\begin{array}{l} \quad \quad\tan A=\sqrt{3}\\\\ \quad\quad \displaystyle\frac{\text{length of opposite side}}{\text{length of adjacent side}}=\sqrt{3}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=60^{\circ} \end{array}$

   
   (d) $\cot A=1$
   
   
   
   

   $\begin{array}{l} \quad \quad\cot A=1\\\\ \quad\quad \displaystyle\frac{\text{length of adjacent side}}{\text{length of opposite side}}=1\\\\ \therefore \quad \text{the triangle is a}\ 45^{\circ} -45^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=45^{\circ} \end{array}$

   
   (e) $\sec A=\sqrt{2}$
   
   
   
   

   $\begin{array}{l} \quad \quad\sec A=\sqrt{2}\\\\ \quad\quad \displaystyle\frac{\text{length of hypotenuse side}}{\text{length of adjacent side}}=\sqrt{2}\\\\ \therefore \quad \text{the triangle is a}\ 45^{\circ} -45^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=45^{\circ} \end{array}$

   
   (f) $\csc A=2$





$\begin{array}{l} \quad \quad\csc A=2\\\\ \quad\quad \displaystyle\frac{\text{length of hypotenuse side}}{\text{length of opposite side}}=2\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=30^{\circ} \end{array}$


For each of the right triangles $A B C$, find the indicated sides.



2. $\angle A=30^{\circ}, \quad c=30$, find $a$.





$\begin{array}{l} \quad\quad \displaystyle \frac{a}{c}=\sin A\\\\ \quad\quad \displaystyle \frac{a}{{30}}=\sin 30^{\circ} \\\\\ \quad\quad \displaystyle \frac{a}{{30}}=\displaystyle \frac{1}{2}\\\\ \quad\quad a=\displaystyle \frac{1}{2}\times 30=15 \end{array}$



3. $\angle A=60^{\circ}, \quad a=15$, find $b$.





$\begin{array}{l} \quad\quad \displaystyle \frac{b}{a}=\cot A\\\\ \quad\quad \displaystyle \frac{b}{15}=\cot 60^{\circ} \\\\\ \quad\quad \displaystyle \frac{b}{15}=\displaystyle \frac{1}{\sqrt{3}}\\\\ \quad\quad b=\displaystyle \frac{15}{\sqrt{3}}=5\sqrt{3} \end{array}$



4. $\angle B=45^{\circ}, \quad a=16$, find $c$.





$\begin{array}{l} \quad\quad \displaystyle \frac{c}{a}=\csc A\\\\ \quad\quad \displaystyle \frac{c}{16}=\csc 45^{\circ} \\\\\ \quad\quad \displaystyle \frac{c}{16}=\sqrt{2}\\\\ \quad\quad b=16\sqrt{2} \end{array}$



5. $\angle B=30^{\circ}, \quad b=8$, find $c$.





$\begin{array}{l} \quad\quad \displaystyle \frac{c}{b}=\csc B\\\\ \quad\quad \displaystyle \frac{c}{8}=\csc 30^{\circ} \\\\ \quad\quad \displaystyle \frac{c}{8}=2\\\\ \quad\quad c=16 \end{array}$



6. A ladder is placed along a wall such that it upper end is touching the top of the wall. The foot of the ladder is $5\ \text{ft}$ away from the wall and the ladder is making an angle of $60^{\circ}$ with the level of the ground. Find the height of the wall.





$\begin{array}{l}\ \ AC=5\ \text{ft}\\\\\ \ \angle A=60{}^\circ ,\ \angle C=90{}^\circ \\\\\ \ BC=?\\\\\ \ \ \text{In right }\triangle ABC,\\\\\ \ \displaystyle\frac{{BC}}{{AC}}=\tan A\\\\\ \ \displaystyle\frac{{BC}}{5}=\tan 60{}^\circ \\\\\ \ \displaystyle\frac{{BC}}{5}=\sqrt{3}\\\\\ \ BC=5\sqrt{3}\ \text{ft}\end{array}$


Find the numerical value of:


7. $\cot ^{3} 45^{\circ}+4 \sin ^{3} 30^{\circ}$.





$\begin{array}{l}\ \ \ {{\cot }^{3}}45{}^\circ +4{{\sin }^{3}}30{}^\circ \\={{(1)}^{3}}+4{{\left( {\displaystyle\frac{1}{2}} \right)}^{3}}\\={{(1)}^{3}}+4\left( {\displaystyle\frac{1}{8}} \right)\\=1+\displaystyle\frac{1}{2}\\=\displaystyle\frac{3}{2}\end{array}$



8. $\tan 60^{\circ} \cot 30^{\circ}+4 \sec ^{2} 30^{\circ}$





$\begin{array}{l}\ \ \ \tan {60}^{\circ }\cot {30}^{\circ }+4\sec^{2}30^{\circ }\\\\=\sqrt{3}\left( {\sqrt{3}} \right)+4{{\left( 2 \right)}^{2}}\\\\=3+16\\\\=19\end{array}$



9. $\tan ^{2} 45^{\circ}+\sin 30^{\circ}-\cos ^{2} 30^{\circ}+2 \tan ^{2} 60^{\circ}$.





$\begin{array}{l} \ \ \ \tan^{2} 45^{\circ} +\sin 30^{\circ} -\cos^2 30^{\circ} +2\tan^2 {60}^{\circ} \\\\ ={1}^{2}+\displaystyle\frac{1}{2}-\left(\displaystyle\frac{\sqrt{3}}{2} \right)^{2}+2\left(\sqrt{3}\right)\\\\ =1+\displaystyle\frac{1}{2}-\displaystyle\frac{3}{4}+2\sqrt{3}\\\\ =\displaystyle\frac{3+8\sqrt{3}}{4} \end{array}$



10. $\displaystyle\frac{1}{2} \sec ^{2} 30^{\circ}+\csc ^{2} 45^{\circ}-2 \tan ^{2} 30^{\circ}$.





$\begin{array}{l} \ \ \ \displaystyle \frac{1}{2}\sec^{2} 30^{\circ} +\csc^{2} 45^{\circ} -2\tan^{2} 30^{\circ} \\\\ =\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{2\sqrt{3}}{3} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}-2\left( \displaystyle \frac{\sqrt{3}}{3} \right)^{2}\\\\ =\displaystyle \frac{1}{2}\left( \displaystyle \frac{4}{3} \right)+\left( 2 \right)-2\left( \displaystyle \frac{1}{3} \right)\\\\=\displaystyle \frac{2}{3}+2-\displaystyle \frac{2}{3}\\\\ =2 \end{array}$

Trigonometric Ratios of 0°, 180°, 270° and 360°

Trigonometric Ratios of 0° 

From the unit circle we have ,

$ \displaystyle \sin 0°=y=0$

$ \displaystyle \cos 0°=x=1$

Therefore

$ \displaystyle \tan 0°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 0°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 0°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 0°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 90° 

Similarly,

$ \displaystyle \sin 90°=y=1$

$ \displaystyle \cos 90°=x=0$

$ \displaystyle \tan 90°=\frac{y}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \cot 90°=\frac{x}{y}=\frac{0}{1}=0$

$ \displaystyle \sec 90°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 90°=\frac{1}{y}=\frac{1}{1}=1$

Trigonometric Ratios of 180° 

Similarly,

$ \displaystyle \sin 180°=y=0$

$ \displaystyle \cos 180°=x=-1$

$ \displaystyle \tan 180°=\frac{y}{x}=\frac{0}{-1}=0$

$ \displaystyle \cot 180°=\frac{x}{y}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \sec 180°=\frac{1}{x}=\frac{1}{-1}=-1$

$ \displaystyle \operatorname{cosec} 180°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 270° 

Similarly,

$ \displaystyle \sin 270°=y=-1$

$ \displaystyle \cos 270°=x=0$

$ \displaystyle \tan 270°=\frac{y}{x}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \cot 270°=\frac{x}{y}=\frac{0}{-1}=0$

$ \displaystyle \sec 270°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 270°=\frac{1}{y}=\frac{1}{-1}=-1$

Trigonometric Ratios of 360° 

Similarly,

$ \displaystyle \sin 360°=y=0$

$ \displaystyle \cos 360°=x=1$

$ \displaystyle \tan 360°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 360°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 360°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 360°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$


Applet တြင္ undefined အတြက္ ∞ သေကၤတ ကို သံုးထားပါသည္။

Trigonometric Ratios of Special Angles

အနား တစ္ဖက္ $ \displaystyle x$ unit ရွိတဲ့ စတုရနး္ $ \displaystyle ABCD$ ဆိုပါစို႔။

စတုရန္း ျဖစ္ေသာေၾကာင့္ ေထာင့္ျဖတ္မ်ဥ္း $ \displaystyle AC$ က သက္ဆိုင္ရာ ေထာင့္မ်ားကို ထက္၀က္ပိုင္းပါတယ္။ ဒါဆိုရင္ရင္ ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္း ထပ္တူညီ ေထာင့္မွန္ ႀတိဂံႏွစ္ခု ျဖစ္လာပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ ႏွစ္နားညီ တဲ့ ေတာင့္မွန္ ႀတိဂံတစ္ခု  ($ \displaystyle 45°-45°$ right triangle လို႔ ေခၚပါတယ္) ရလာပါတယ္။ $ \displaystyle AB=BC=x$ ျဖစ္တာ ေၾကာင့္ $ \displaystyle AC$ ရဲ့ အလ်ားကို Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

        $ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ =2{{x}^{2}}\\\\\therefore \ \ AC=\sqrt{2}x\end{array}$

ဒါဆိုရင္ $ \displaystyle AB:BC:AC=1:1:\sqrt{2}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 45°-45°$ right triangle ကို special triangle လို႔ ေခၚၿပီး $ \displaystyle 45°$ ေထာင့္ကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

အနားေတြရဲ့ အခ်ိဳးကို သိၿပီဆိုေတာ့ $ \displaystyle 45°$ angle ရဲ့ trigonometric ratios ေတြကို အလြယ္ တကူရွာႏိုင္ၿပီေပါ့။

Trigonometric Ratios of $ \displaystyle 45°$

$ \displaystyle \sin 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \tan 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \cot 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \sec 45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

$ \displaystyle \operatorname{cosec}45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

ယခုတစ္ခါ အနားတစ္ဘက္ $ \displaystyle 2x$ unit ရွိတဲ့ သံုးနားညီႀတိဂံတစ္ခု $ \displaystyle ABD$ ကို စဥ္းစားၾကည့္မယ္။ ေထာင့္စြန္းမွတ္ $ \displaystyle A$ မွ $ \displaystyle BD$ ေပၚသို႔ အျမင့္မ်ဥ္း $ \displaystyle AC$ ကိုဆြဲလိုက္မယ္ ဆိုရင္ သံုးနားညီ ႀတိဂံျဖစ္တာေၾကာင့္ $ \displaystyle AC$ ဟာ အလယ္မ်ဥ္းလည္း ျဖစ္သလို ေထာင့္ထက္၀က္ပိုင္း မ်ဥ္းလဲ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္ ပံုမွာ ျပထားတဲ့ အတိုင္း $ \displaystyle AC$ ဟာ $ \displaystyle \vartriangle ABD$ ကို ထပ္တူညီ ႀတိဂံ ႏွစ္ခုအျဖစ္ ပိုင္းျဖတ္ လိုက္ပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ $ \displaystyle 30°-60°$ right triangle ျဖစ္ၿပီး $ \displaystyle AC$ ရဲ့ အလ်ားကိုေတာ့ Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

$ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}-B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}-{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =3{{x}^{2}}\\\\\therefore \ \ \ \ AC=\sqrt{3}x\end{array}$

ဒါဆိုရင္ $ \displaystyle BC:AB:AC=1:2:\sqrt{3}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 30°-60°$ right triangle ကိုလည္း special triangle လို႔ ေခၚၿပီး $ \displaystyle 30°$ နဲ႔ $ \displaystyle 60°$ ေထာင့္ ေတြကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

Trigonometric Ratios of $ \displaystyle 30°$

$ \displaystyle \sin 30{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \cos 30{}^\circ =\frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \tan 30{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \cot 30{}^\circ =\frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \sec 30{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \operatorname{cosec}30{}^\circ =\frac{{2x}}{x}=2$

Trigonometric Ratios of $ \displaystyle 60°$

$ \displaystyle \sin 60{}^\circ = \frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \cos 60{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \tan 60{}^\circ = \frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \cot 60{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \sec 60{}^\circ = \frac{{2x}}{x}=2$

$ \displaystyle \operatorname{cosec}60{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$