# Consider the non-right triangle shown below that has lengths of 1.6, 1.751 and 2

Consider the non-right triangle shown below that has lengths of 1.6, 1.751 and 2.6 cm and interior angle measures of 0.72, $$\displaystyle\alpha$$, and $$\displaystyle\beta$$ degrees.

a) What is the value of $$\displaystyle\alpha$$?
b) What is the value of $$\displaystyle\beta$$?

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Step 1
Let,
$$\displaystyle{a}={1.6}{c}{m}$$
$$\displaystyle{b}={2.6}{c}{m}$$
$$\displaystyle{c}={1.751}{c}{m}$$
And,
$$\displaystyle{A}=\alpha,\ {B}=\beta,\ {C}={41.253}^{{\circ}}$$
According to sine law:
$$\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}={\frac{{{\sin{{C}}}}}{{{c}}}}$$
Step 2
Thus,
$$\displaystyle{\frac{{{\sin{\alpha}}}}{{{1.6}}}}={\frac{{{\sin{{41.253}}}^{{\circ}}}}{}}$$
$$\displaystyle{\sin{\alpha}}={\frac{{{1.6}\times{\sin{{41.253}}}^{{\circ}}}}{{{1.751}}}}$$
$$\displaystyle{\sin{\alpha}}={0.60252}$$
$$\displaystyle\therefore\alpha={37.051}^{{\circ}}$$
Now, using triangle property
$$\displaystyle{A}+{B}+{C}={180}^{{\circ}}$$
$$\displaystyle\alpha+\beta+{41.253}={180}^{{\circ}}$$
$$\displaystyle{37.051}+\beta={180}^{{\circ}}-{41.253}^{{\circ}}$$
$$\displaystyle\beta={138.747}-{37.051}$$
$$\displaystyle\therefore\beta={101.696}^{{\circ}}$$
Step 3
Therefore,
$$\displaystyle\alpha={37.051}^{{\circ}}$$
$$\displaystyle\beta={101.696}^{{\circ}}$$