In $\triangle$ $ABC,$ right-angled at $B$, $AB =5\, cm$ and $\angle ACB =30^{\circ}$ (see $Fig.$). Determine the lengths of the sides $BC$ and $AC .$
To find the length of the side $BC ,$ we will choose the trigonometric ratio involving $BC$ and the given side $AB$. since $BC$ is the side adjacent to angle $C$ and $AB$ is the side opposite to angle $C ,$ therefore
$\frac{ AB }{ BC }=\tan C$
$\frac{5}{ BC }=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
which gives $BC =5 \sqrt{3} \,cm$
To find the length of the side $AC ,$ we consider
$\sin 30^{\circ}=\frac{ AB }{ AC }$
$\frac{1}{2}=\frac{5}{ AC }$
$AC =10 \,cm$
Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
$AC =\sqrt{ AB ^{2}+ BC ^{2}}=\sqrt{5^{2}+(5 \sqrt{3})^{2}} cm =10 \,cm$
In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
If $\sin ( A - B )=\frac{1}{2}, \cos ( A + B )=\frac{1}{2}, 0^{\circ} < A + B \leq 90^{\circ}, A > B ,$ find $A$ and $B$
State whether the following are true or false. Justify your answer.
$(i)$ $\cos A$ is the abbreviation used for the cosecant of angle $A$
$(ii)$ cot $A$ is the product of cot and $A$.
$(iii)$ $\sin \theta=\frac{4}{3}$ for some angle $\theta$.
If $\sec 4 A =\operatorname{cosec}\left( A -20^{\circ}\right),$ where $4 A$ is an acute angle, find the value of $A$. (in $^{\circ}$)
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$