Notes: `A = 36^@, a = 8, b = 5` Use the law of sines to solve the triangle. (Find missing sides/angles) Round answers to 2 decimal places.

Friday, 12 September 2014

$\displaystyle{A}={{36}}^{\circ},{a}={8},{b}={5}$ Use the law of sines to solve the triangle. (Find missing sides/angles) Round answers to 2 decimal places.

The sine law indicates that in a triangle;

$\displaystyle\frac{{{\sin{{A}}}}}{{a}}=\frac{{{\sin{{B}}}}}{{b}}=\frac{{{\sin{{C}}}}}{{c}}$

a,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image.

$\displaystyle\frac{{{\sin{{A}}}}}{{a}}=\frac{{{\sin{{B}}}}}{{b}}$

$\displaystyle\frac{{\sin{{36}}}}{{8}}=\frac{{\sin{{B}}}}{{5}}$

$\displaystyle{B}={{\sin}}^{{-{1}}}{\left({\sin{{36}}}\times\frac{{5}}{{8}}\right)}={{21.55}}^{{0}}$

Angle C $\displaystyle={180}-{21.55}-{36}={{122.45}}^{{0}}$

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

$\displaystyle{c}=\frac{{{\sin{{122.45}}}\times{5}}}{{\sin{{21.55}}}}={11.49}$

So the answers are;

Angle B = 21.55 deg

Angle C = 122.45 deg

Length c = 11.49 units

...

The sine law indicates that in a triangle;

$\displaystyle\frac{{{\sin{{A}}}}}{{a}}=\frac{{{\sin{{B}}}}}{{b}}=\frac{{{\sin{{C}}}}}{{c}}$

a,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image.

$\displaystyle\frac{{{\sin{{A}}}}}{{a}}=\frac{{{\sin{{B}}}}}{{b}}$

$\displaystyle\frac{{\sin{{36}}}}{{8}}=\frac{{\sin{{B}}}}{{5}}$

$\displaystyle{B}={{\sin}}^{{-{1}}}{\left({\sin{{36}}}\times\frac{{5}}{{8}}\right)}={{21.55}}^{{0}}$

Angle C $\displaystyle={180}-{21.55}-{36}={{122.45}}^{{0}}$

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

$\displaystyle{c}=\frac{{{\sin{{122.45}}}\times{5}}}{{\sin{{21.55}}}}={11.49}$

So the answers are;

Angle B = 21.55 deg

Angle C = 122.45 deg

Length c = 11.49 units

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