computing $\alpha \backslash alpha$

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

$\alpha ={30.7}^{\circ }\backslash alpha=30.7^\{\backslash circ\}$

computing $\beta \backslash beta$

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

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

$\beta ={59.3}^{\circ }\backslash beta=59.3^\{\backslash circ\}$

computing $bb$

Then you compute $bb$ using the Pythagorean theorem.

${\left(24.9\mathrm{c}\mathrm{m}\right)}^2={\left(12.7\mathrm{c}\mathrm{m}\right)}^2+b^2\backslash left(24\{.\}9\backslash ,\backslash mathrm\{cm\}\backslash right)^2=\backslash left(12\{.\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2+b^2$

$\mathrm{\mid }-{\left(12.7\mathrm{c}\mathrm{m}\right)}^2|\{\}-\backslash left(12\{.\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2={\left(24.9\mathrm{c}\mathrm{m}\right)}^2-{\left(12.7\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(24\{.\}9\backslash ,\backslash mathrm\{cm\}\backslash right)^2-\backslash left(12\{.\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2=458.72{\mathrm{c}\mathrm{m}}^{2}b^2=458\{.\}72\backslash ,\backslash mathrm\{cm\}^2$

Take the root.

$b\approx 21.4\mathrm{c}\mathrm{m}b\backslash approx21\{.\}4\backslash ,\backslash mathrm\{cm\}$

computing $\beta \backslash beta$

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

$\beta ={40.6}^{\circ }\backslash beta=40.6^\{\backslash circ\}$

computing $\gamma \backslash gamma$

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

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

computing $cc$

Then you compute $cc$ using the Pythagorean theorem.

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

$(645\mathrm{m}{)}^2=(420\mathrm{m}{)}^2+c^2(645\backslash ,\backslash mathrm\; m)^2=(420\backslash ,\backslash mathrm\; m)^2+c^2$

$c^2=(645\mathrm{m}{)}^2-(420\mathrm{m}{)}^{2}c^2=\; (645\backslash ,\backslash mathrm\; m)^2\; -\; (420\backslash ,\backslash mathrm\; m)^2$

$c^2=239625{\mathrm{m}}^{2}c^2=239\backslash ,625\backslash ,\backslash mathrm\; m^2$

Take the root.

$c\approx 490\mathrm{m}c\backslash approx490\backslash ,\backslash mathrm\; m$

computing $bb$

Compute $bb$ using the Pythagorean theorem.

$b^2={\left(30.7\mathrm{c}\mathrm{m}\right)}^2+{\left(15.8\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(30\{.\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2+\backslash left(15\{.\}8\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2=1192.13{\mathrm{c}\mathrm{m}}^2\mathrm{\mid }\sqrt{}b^2=1192\{.\}13\backslash ,\backslash mathrm\{cm\}^2\; \backslash ;\backslash ;\backslash ;\backslash ;|\backslash sqrt\{\}$

$b\approx 34.5\mathrm{c}\mathrm{m}b\backslash approx34\{.\}5\backslash ,\backslash mathrm\{cm\}$

computing $\alpha \backslash alpha$

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

$\tan \left(\alpha \right)={\displaystyle \frac{30.7\mathrm{c}\mathrm{m}}{15.8\mathrm{c}\mathrm{m}}}\backslash tan\backslash left(\backslash alpha\backslash right)=\backslash dfrac\{30\{.\}7\backslash ,\backslash mathrm\{cm\}\}\{15\{.\}8\backslash ,\backslash mathrm\{cm\}\}$

$\alpha \approx {62.7}^{\circ }\backslash alpha\backslash approx62.7^\{\backslash circ\}$

computing $\gamma \backslash gamma$

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

$\gamma \approx {180}^{\circ }-{90}^{\circ }-{62.7}^{\circ }\backslash gamma\backslash approx180^\{\backslash circ\}-90^\{\backslash circ\}-62.7^\{\backslash circ\}$

$\gamma \approx {27.3}^{\circ }\backslash gamma\backslash approx27.3^\{\backslash circ\}$

computing $\beta \backslash beta$

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

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

$\beta ={55}^{\circ }\backslash beta=55^\{\backslash circ\}$

computing $aa$

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

$a=\sin \left({35}^{\circ }\right)\cdot 12.5\mathrm{c}\mathrm{m}a=\backslash sin\backslash left(35^\backslash circ\backslash right)\backslash cdot12\{.\}5\backslash ,\backslash mathrm\{cm\}$

$a=7.2\mathrm{c}\mathrm{m}a=7\{.\}2\backslash ,\backslash mathrm\{cm\}$

computing $bb$

Then you compute $bb$ using the Pythagorean theorem.

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

${\left(12.5\mathrm{c}\mathrm{m}\right)}^2={\left(7.2\mathrm{c}\mathrm{m}\right)}^2+b^2\mathrm{\mid }-{\left(7.2\mathrm{c}\mathrm{m}\right)}^2\backslash left(12\{.\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2=\backslash left(7\{.\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2+b^2\; \backslash :\backslash :\backslash ;\backslash ;\backslash ;\backslash ;|\{\}-\backslash left(7\{.\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2={\left(12.5\mathrm{c}\mathrm{m}\right)}^2-{\left(7.2\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(12\{.\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2-\backslash left(7\{.\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2\approx 104.4{\mathrm{c}\mathrm{m}}^{2}b^2\backslash approx104\{.\}4\backslash ,\backslash mathrm\{cm\}^2$

Take the root.

$b\approx 10.2\mathrm{c}\mathrm{m}b\backslash approx10\{.\}2\backslash ,\backslash mathrm\{cm\}$

computing $bb$ (alternative solution with the cosine)

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

Solve for $bb$

$b=\cos \left(\alpha \right)\cdot c\approx 10.2\mathrm{c}\mathrm{m}b\; =\; \backslash cos\backslash left(\; \backslash alpha\; \backslash right)\; \backslash cdot\; c\; \backslash approx\; 10\{.\}2\backslash ,\backslash mathrm\{cm\}$

computing $\beta \backslash beta$

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

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

$\beta ={49.7}^{\circ }\backslash beta=49.7^\{\backslash circ\}$

computing $bb$

$\sin \left({49.7}^{\circ }\right)=\frac{b}{10.5\mathrm{c}\mathrm{m}}\mathrm{\mid }\cdot 10.5\mathrm{c}\mathrm{m}\backslash sin\backslash left(49\{.\}7^\backslash circ\backslash right)=\backslash frac\{b\}\{10.5\backslash ,\backslash mathrm\{cm\}\}\backslash :\backslash :\backslash ;\backslash ;\backslash ;\backslash ;\; |\{\}\backslash cdot10\{.\}5\backslash ,\backslash mathrm\{cm\}$

$b=\sin \left({49.7}^{\circ }\right)\cdot 10.5\mathrm{c}\mathrm{m}b=\backslash sin\backslash left(49\{.\}7^\backslash circ\backslash right)\backslash cdot10\{.\}5\backslash ,\backslash mathrm\{cm\}$

$b\approx 8\mathrm{c}\mathrm{m}b\backslash approx8\backslash ,\backslash mathrm\{cm\}$

computing $cc$

Then you compute $cc$ using the Pythagorean theorem.

${\left(10.5\mathrm{c}\mathrm{m}\right)}^2=c^2+{\left(8\mathrm{c}\mathrm{m}\right)}^2\backslash left(10\{.\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2=c^2+\backslash left(8\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$c^2=46.25{\mathrm{c}\mathrm{m}}^{2}c^2=46\{.\}25\backslash ,\backslash mathrm\{cm\}^2$

Take the root.

$c\approx 6.8\mathrm{c}\mathrm{m}c\backslash approx6\{.\}8\backslash ,\backslash mathrm\{cm\}$