How do you solve the triangle given C=15^circ15', a=6.25, b=2.15?

Jan 29, 2017

Covert minutes to decimal degrees
Use The Law of Cosines to find the length of side c
Use The Law of Sines to find $B$
Use the sum of the interior angles equals ${180}^{\circ}$ to find $A$

Explanation:

Use the factor ${1}^{\circ} / \left(60 '\right)$ to convert the minutes to decimal degrees:

$\frac{15 '}{1} \frac{{1}^{\circ}}{60 '} = {\left(\frac{15}{60}\right)}^{\circ} = {0.25}^{\circ}$

$C = {15.25}^{\circ}$

Use The Law of Cosines to find the length of side c:

c=sqrt(a^2+b^2-2(a)(b)cos(C)

c=sqrt((6.25)^2+(2.15)^2-2(6.25)(2.15)cos(15.25^@)

$c \approx 4.21$

Use The Law of Sines to find B:

$\sin \frac{B}{b} = \sin \frac{C}{c}$

$B = {\sin}^{-} 1 \left(\sin \left(C\right) \frac{b}{c}\right)$

$B = {\sin}^{-} 1 \left(\sin \left({15.25}^{\circ}\right) \frac{2.15}{4.21}\right)$

$B \approx {7.72}^{\circ}$

To find A use the some if the interior angle equal ${180}^{\circ}$:

$A + B + C = {180}^{\circ}$

$A + {7.72}^{\circ} + {15.25}^{\circ} = {180}^{\circ}$

$A = {157.03}^{\circ}$