A triangle has sides of lengths 9, 10, and 11. Then the number of degrees in the smallest angle of this triangle, correct to the nearest tenth of a degree, will be
is it 39.8o ???
2006-09-01 10:32:13 · 5 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
50°28'44" or 50.5°
2006-09-01 10:39:46 · answer #1 · answered by bretttwarwick 3 · 0⤊ 0⤋
1. The smallest angle is opposite to the smallest side.
2. Use cosine rule
9^2 = 10^2 + 11^2 -2 x 10 x 11 x cosA
cosA = 0.636363...
A = 50 degrees. So the answer is No.
2006-09-01 17:40:57 · answer #2 · answered by Thermo 6 · 0⤊ 0⤋
well we have this formula
if a , b & c are the tringle's sides and a =9 , b=10 & c= 11;so we have ;
a^2 = b^2 + c^2 -2 b c ( cos (a))
{ if we just have one angle it would be very easy to find the other ones }
(9^2) = (10^2) + (11^2 )-(2 * 10 * 11* cos(a))
81 = 100 - 121 - 110*cos(a)
cos(a) =-140 / -220
cos(a) = 0.6363... a bar over ' 63 '
a = 50â¦
good question. good luck.
2006-09-01 18:16:24 · answer #3 · answered by sweetie 5 · 0⤊ 0⤋
thermo is right, but I get 50.4788 rounded to 50.5 degrees
2006-09-01 18:23:21 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Nope, it's 50.5°. Check your math.
2006-09-01 17:40:07 · answer #5 · answered by Pascal 7 · 1⤊ 0⤋