AB = 24 cm and BC = 7 cm
tan C = Opposite side/Adjacent side
tan C = $\dfrac{24}{7}$
sin 30° = $\dfrac{1}{2}$ sin 60° =$\dfrac{\sqrt{3}}{2}$, cos 30° = $\dfrac{\sqrt{3}}{2}$ and cos 60°= $\dfrac{1}{2}$
Putting these values, we get:
$(\dfrac{1}{2}+\dfrac{1}{2})-(\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2})$
=$ 1 – [\dfrac{(2\sqrt{3})}{2}]$
= 1 – $\sqrt{3}$
tan 60° =$ \sqrt{3}$ and cot 30° = $\sqrt{3}$
Hence, $\dfrac{tan 60°}{cot 30°}$ =$\dfrac{ \sqrt{3}}{\sqrt{3}}$= 1
We know, by trigonometry identities,
$sin^{2}A$ + $cos^{2}A$ = 1
1 –$cos^{2}A$ =$sin^{2}A$
By trigonometry identities.
Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}
By trigonometry identities, we know:
$1 + tan^{2}X $= $sec^{2}X$
And sec X = $\dfrac{1}{cos X}$ =$ \dfrac{1}{(\dfrac{2}{3})}$ =$ \dfrac{3}{2}$
Hence,
$1 + tan^{2}X$ = $(\dfrac{3}{2})^{2}$ =$ \dfrac{9}{4}$
$tan^{2}X$ = $(\dfrac{9}{4}) $– 1 = $\dfrac{5}{4}$
tan X = $\dfrac{\sqrt{5}}{2}$
cos X = $\dfrac{a}{b}$
By trigonometry identities, we know that:
$sin^{2}X + cos^{2}X$ = 1
$sin^{2}X$ = $1 – cos^{2}X$ = $1-(a/b)^{2}$
sin X = $\dfrac{\sqrt{(b^{2}-a^{2})}}{b}$
sin 60° = $\dfrac{\sqrt{3}}{2}$, sin 30° =$\dfrac{1}{2}$, cos 60° =$\dfrac{1}{2}$ and cos 30° = $\dfrac{\sqrt{3}}{2}$
Therefore,
($\dfrac{\sqrt{3}}{2}$) x ($\dfrac{\sqrt{3}}{2}$) + ($\dfrac{1}{2}$) x ($\dfrac{1}{2}$)
= ($\dfrac{3}{4}$) + ($\dfrac{1}{4}$)
= $\dfrac{4}{4}$
= 1
tan 30° = $\dfrac{1}{\sqrt{3}}$
Putting this value we get;
$\dfrac{[2(\dfrac{1}{\sqrt{3}})]}{[1 + (\dfrac{1}{\sqrt{3}})^{2}] }$
=$\dfrac{(\dfrac{2}{\sqrt{3}})}{( \dfrac{4}{3}) }$
= $\dfrac{6}{4\sqrt{3} }$
=$\dfrac{\sqrt{3}}{2 }$
= sin 60°
sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 0