Answer:
D. m∠A=43, m∠B=55, a=20
Step-by-step explanation:
Given:
∆ABC,
m<C = 82°
AB = c = 29
AC = b = 24
Required:
m<A, m<C, and a (BC)
SOLUTION:
Find m<B using the law of sines:
[tex] \frac{sin(B)}{b} = \frac{sin(C)}{c} [/tex]
[tex] \frac{sin(B)}{24} = \frac{sin(82)}{29} [/tex]
[tex] sin(B)*29 = sin(82)*24 [/tex]
[tex] \frac{sin(B)*29}{29} = \frac{sin(82)*24}{29} [/tex]
[tex] sin(B) = \frac{sin(82)*24}{29} [/tex]
[tex] sin(B) = 0.8195 [/tex]
[tex] B = sin^{-1}(0.8195) [/tex]
[tex] B = 55.0 [/tex]
m<B = 55°
Find m<A:
m<A = 180 - (82 + 55) => sum of angles in a triangle.
= 180 - 137
m<A = 43°
Find a using the law of sines:
[tex] \frac{a}{sin(A)} = \frac{b}{sin(B)} [/tex]
[tex] \frac{a}{sin(43)43} = \frac{24}{sin(55)} [/tex]
Cross multiply
[tex] a*sin(55) = 25*sin(43) [/tex]
[tex] a = \frac{25*sin(43)}{sin(53)} [/tex]
[tex] a = 20 [/tex] (approximated)
