# How do you solve the triangle given A=40, b=7, c=6?

Jan 13, 2017

angle B= 81°56'14'', a = 4.544, angle C=58°03'46''

#### Explanation:

Cosine rule:

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c C o s A$

a^2=(7^2)+(6^2)-2 xx 7 xx 6 xx Cos40°

${a}^{2} = 49 + 36 - 84 \times 0.766044443$

${a}^{2} = 85 - 64.34773322$

${a}^{2} = 20.65226678$

$a = \sqrt{20.65226678}$

$a = 4.544 = B C$

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c C o s B$

${b}^{2} = {\left(4.544\right)}^{2} + {6}^{2} - 2 \times 4.544 \times C o s B$

${7}^{2} = {\left(4.544\right)}^{2} + 36 - 54.528 \times C o s B$

${7}^{2} = 20.648 + 36 - 54.528 \times C o s B$

${7}^{2} = 56.648 - 54.528 \times C o s B$

$- 54.528 \cos B = 49 - 56.648$

$54.528 C o s B = 56.648 - 49$

$C o s B = \frac{7.648}{54.528}$

$C o s B = 0.140258216$

angleB=81° 56’ 14’’

180°-(40°+81° 56' 14'')=58° 03' 46'' =angle C

Check:

(BC)/(Sin40)=6/(Sin58° 03’ 46’’)

BC=(6 xx Sin40°)/(Sin58° 03’ 46’’)

$B C = \frac{3.856725658}{0.848628208}$

$B C = 4.545 = a$