In a right $△\mathrm{ABC}$, right-angled at C, if $\angle \mathrm{B}={60}^{\circ }$ and AB = 15 units. Find the remaining angles and sides.

We can solve this using angle sum property.

The angle sum property of a triangle states that the sum of the interior angles of any triangle is always $180^{\circ }$.

Given, right angle at C in the triangle $\mathrm{ABC}$. 

In this triangle, $\angle \mathrm{B}={60}^{\circ }$, $\angle \mathrm{C}={90}^{\circ }$ and AB = 15 

Now, in triangle ABC, 

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}={180}^{\circ }$         [Sum of the all angles in a triangle = 180°] 

${60}^{\circ }+\angle \mathrm{A}+{90}^{\circ }={180}^{\circ }$

$\angle A={30}^{\circ }$

Now from the trigonometry property, 

$\mathrm{sinA}=\frac{\mathrm{CB}}{\mathrm{AB}}$

$\sin {30}^{\circ }=\frac{\mathrm{CB}}{15}$
$\frac{1}{2}\times 15=\mathrm{CB}$    [Since, $\sin {30}^{\circ }=\frac{1}{2}$]

$\mathrm{CB}=7.5 \mathrm{units}$ 

Now,

${\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2$   [By using of Pythagoras theorem]

${15}^2={\mathrm{AC}}^2+{\left(\frac{15}{2}\right)}^2$

$\mathrm{AC}=225-\left(\frac{225}{4}\right)$

${\mathrm{AC}}^2=225\left(\frac{3}{4}\right)$

$\mathrm{AC}=\frac{15\sqrt{3}}{2}\mathrm{ }\mathrm{units}$