In triangle ABC, AB= 10cm, BC= a√3 cm, AC=5√13 cm and angle ABC=150 degrees.
Calculate:
a) the value of a
b) the exact area of triangle ABC
this is the question that i got 8.66 for, but the textbook answer says it is 5.
howww?
thankss
xxx
In triangle ABC, AB= 10cm, BC= a√3 cm, AC=5√13 cm and angle ABC=150 degrees.
Calculate:
a) the value of a
b) the exact area of triangle ABC
this is the question that i got 8.66 for, but the textbook answer says it is 5.
howww?
thankss
xxx
Start with the cosine rule. AC^2 = AB^2 + BC^2 - 2AB.BC.cos ABC. Here we have 325 = 100 + 3a^2 - 2.10.a sqrt(3).cos(150). Now cos(150) = -cos(30) = -sqrt(3)/2, so 325 = 100 + 3a^2 + 30a. So a^2 + 10a - 75 = 0, with roots a=5 or -15. The positive solution is the only possible one, so a = 5 and the sides are 10, 5sqrt(3) and 5sqrt(13).