computing $\alpha$

$\sin\left(\alpha\right)=\frac{12.7\,\mathrm{cm}}{24.9\,\mathrm{cm}}$

$\alpha=30.7^{\circ}$

computing $\beta$

You can calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$.

$\beta=180^{\circ}-90^{\circ}-30.7^{\circ}$

$\beta=59.3^{\circ}$

computing $b$

Then you compute $b$ using the Pythagorean theorem.

$\left(24{.}9\,\mathrm{cm}\right)^2=\left(12{.}7\,\mathrm{cm}\right)^2+b^2$

$|{}-\left(12{.}7\,\mathrm{cm}\right)^2$

$b^2=\left(24{.}9\,\mathrm{cm}\right)^2-\left(12{.}7\,\mathrm{cm}\right)^2$

$b^2=458{.}72\,\mathrm{cm}^2$

Take the root.

$b\approx21{.}4\,\mathrm{cm}$

computing $\beta$

$\sin\left(\beta\right)=\frac{420\,\mathrm m}{645\,\mathrm m}$

$\beta=40.6^{\circ}$

computing $\gamma$

You can calculate $\gamma$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$.

$\gamma=180^{\circ}-90^{\circ}-40.6^{\circ}=49.4^{\circ}$

computing $c$

Then you compute $c$ using the Pythagorean theorem.

Note in particular that c is not the hypotenuse of the triangle, but $a$ (see picture above), so that the Pythagorean form remains similar, but now $a$, $b$ and $c$ do not take the familiar roles, where $a$,$b$ would be the cathetuses and $c$ the hypotenuse.

$(645\,\mathrm m)^2=(420\,\mathrm m)^2+c^2$

$c^2= (645\,\mathrm m)^2 - (420\,\mathrm m)^2$

$c^2=239\,625\,\mathrm m^2$

Take the root.

$c\approx490\,\mathrm m$

computing $b$

Compute $b$ using the Pythagorean theorem.

$b^2=\left(30{.}7\,\mathrm{cm}\right)^2+\left(15{.}8\,\mathrm{cm}\right)^2$

$b^2=1192{.}13\,\mathrm{cm}^2 \;\;\;\;|\sqrt{}$

$b\approx34{.}5\,\mathrm{cm}$

computing $\alpha$

$\tan\left(\alpha\right)=\frac{a}{c}$

$\tan\left(\alpha\right)=\dfrac{30{.}7\,\mathrm{cm}}{15{.}8\,\mathrm{cm}}$

$\alpha\approx62.7^{\circ}$

computing $\gamma$

You may calculate $\gamma$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$.

$\gamma\approx180^{\circ}-90^{\circ}-62.7^{\circ}$

$\gamma\approx27.3^{\circ}$

computing $\beta$

You may calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$.

$\beta=180^{\circ}-90^{\circ}-35^{\circ}$

$\beta=55^{\circ}$

computing $a$

$\sin\left(35^\circ\right)=\frac{a}{12.5\,\mathrm{cm}} \:\:\;\;\;\;|{}\cdot12{.}5\,\mathrm{cm}$

$a=\sin\left(35^\circ\right)\cdot12{.}5\,\mathrm{cm}$

$a=7{.}2\,\mathrm{cm}$

computing $b$

Then you compute $b$ using the Pythagorean theorem.

$c^2=a^2+b^2$

$\left(12{.}5\,\mathrm{cm}\right)^2=\left(7{.}2\,\mathrm{cm}\right)^2+b^2 \:\:\;\;\;\;|{}-\left(7{.}2\,\mathrm{cm}\right)^2$

$b^2=\left(12{.}5\,\mathrm{cm}\right)^2-\left(7{.}2\,\mathrm{cm}\right)^2$

$b^2\approx104{.}4\,\mathrm{cm}^2$

Take the root.

$b\approx10{.}2\,\mathrm{cm}$

computing $b$ (alternative solution with the cosine)

$\cos\left(\alpha\right)=\frac{b}{c}$

Solve for $b$

$b = \cos\left( \alpha \right) \cdot c \approx 10{.}2\,\mathrm{cm}$

computing $\beta$

You may calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$.

$\beta=180^{\circ}-90^{\circ}-40.3^{\circ}$

$\beta=49.7^{\circ}$

computing $b$

$\sin\left(49{.}7^\circ\right)=\frac{b}{10.5\,\mathrm{cm}}\:\:\;\;\;\; |{}\cdot10{.}5\,\mathrm{cm}$

$b=\sin\left(49{.}7^\circ\right)\cdot10{.}5\,\mathrm{cm}$

$b\approx8\,\mathrm{cm}$

computing $c$

Then you compute $c$ using the Pythagorean theorem.

$\left(10{.}5\,\mathrm{cm}\right)^2=c^2+\left(8\,\mathrm{cm}\right)^2$

$c^2=46{.}25\,\mathrm{cm}^2$

Take the root.

$c\approx6{.}8\,\mathrm{cm}$