Question
Applications of trigonometric functions
Solve the missing parts of the triangle satisfying
a=125
b=150
alpha=55 degrees
There are two solutions
Solve the missing parts of the triangle satisfying
a=125
b=150
alpha=55 degrees
There are two solutions
Answers
Steve
a/sin alpha = b/sin beta
125/sin 55 = 150/sin beta
sin beta = .819/125 * 150 = .9828
beta = 79.25 deg or 100.75 deg
gamma = 180 - (alpha+beta)
= 45.75 or 24.25
sin gamma = .716 or .411
c / sin gamma = 152.6
c = 152.6*.716 or 152.6*.411
= 109.26 or 62.72
125/sin 55 = 150/sin beta
sin beta = .819/125 * 150 = .9828
beta = 79.25 deg or 100.75 deg
gamma = 180 - (alpha+beta)
= 45.75 or 24.25
sin gamma = .716 or .411
c / sin gamma = 152.6
c = 152.6*.716 or 152.6*.411
= 109.26 or 62.72