標題:
Problem about triangle
發問:
In triangle ABC, AB=3, BC=4, AC=5.O is a point inside triangle ABC. If angle OAB=angle OBC= angle OCA=x Find the value of tan x,length of AO,BO and CO. 更新: detail steps are required
最佳解答:
(1)By Pythagoras thm, triangle ABC is a right-angled triangle with angle ABC = 90. and angle AOB = 90- x. Therefore, in triangle ABO, angle AOB = 180 - x -(90 -x ) = 90. (2) angle ACB = arctan(3/4) = 36.9, therefore, angle OCB = 36.9 - x. Therefore, in triangle OBC, angle COB = 180 - x -(36.9 - x) = 143.1 (3) Angle AOC = 360 - 90 - 143.1 = 126.9 (4)Let OA = p, OB = r and OC = s. Apply sine rule to triangle OBC, we get 4/sin143.1 = s/sinx................(a) Apply sine rule to triangle AOC, we get 5/sin126.9 = p/sinx................(b) Apply sine rule to triangle AOB, we get 3/sin90 = r/sinx......................(c) (5) (c)/(b), (3/sin90)/(5/sin126.9) = (r/sinx)/(p/sinx) = r/p = tanx. Therefore, tanx = 3sin126.9/5 = 0.48. And x = arctan(0.48) = 25.6. (6)From (a), s = OC = 4sin25.6/sin143.1 = 2.88. From (b), p = OA = 5sin25.6/sin126.9 = 2.70. From (c), r = OB = 3sin25.6 = 1.30 2008-06-18 19:26:46 補充： r/p = tanx because r/p = OB/OA and angle BOA = 90.
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