Use the information given in the following figure to evaluate :

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}$.

From the figure, in Δ ADC,

⇒ AC² = DC² + AD² (∵ AC is hypotenuse)

⇒ 20² = DC² + 12²

⇒ 400 = DC² + 144

⇒ DC² = 400 - 144

⇒ DC² = 256

⇒ DC = $\sqrt{256}$

⇒ DC = 16

BD = BC - DC

= 21 - 16 = 5

In Δ ABD,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = 12² + 5²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$

⇒ AB = 13

sin x = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BD}{AB} = \dfrac{5}{13}$

sin y = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AC} = \dfrac{12}{20} = \dfrac{3}{5}$

cot y = $\dfrac{Base}{Perpendicular}$

$= \dfrac{DC}{AD} = \dfrac{16}{12} = \dfrac{4}{3}$

Now,

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}\\[1em] = \dfrac{10}{\dfrac{5}{13}} + \dfrac{6}{\dfrac{3}{5}} - 6 \times \dfrac{4}{3}\\[1em] = \dfrac{10 \times 13}{5} + \dfrac{6 \times 5}{3} - \dfrac{24}{3}\\[1em] = \dfrac{130}{5} + \dfrac{30}{3} - \dfrac{24}{3}\\[1em] = 26 + 10 - 8\\[1em] = 28$

Hence, $\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y} = 28.$