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) $\csc A=2$

  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.

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

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

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

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

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

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

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

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

  11. 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.

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

    Find the numerical value of:

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

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

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

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

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

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

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

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