Find the largest angle of a triangle with sides 6, 8 and 9
2007-05-12 06:24:02 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You posted ALL your homework? If you don't do it yourself, how will you pass your test?
You can use the sine rule:
sinA / a = sinB / b = sinC / c.
The largest angle will be opposite the largest side.
Now go do it yourself!
2007-05-12 06:33:06 · answer #1 · answered by Marie Antoinette 5 · 0⤊ 0⤋
You automatically know that the largest angle will be the one opposite from the side with the length of 9.
The side with the length of 9 is the longest side, so the angle opposite of that will be the largest angle.
2007-05-12 13:31:00 · answer #2 · answered by Kayla Arielle M. 4 · 0⤊ 0⤋
9² = 6² + 8² - 96 cos a
81 = 100 - 96 cos a
96 cos a = 19
cos a = 19 / 96
a = 78.6 °
2007-05-13 06:59:50 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Cosine rule-
a2= b2 + c2 - 2bc cos A
81 = 36 + 64 - 2*8*6 cos A
81 = 100 - (96 cos A)
19 = 96 cos A
cos A = 19 / 96
A = inverse cos (19/96)
A= 78. 6degrees
180- 78.6 = 101.4 degrees
2007-05-12 16:15:09 · answer #4 · answered by Anonymous · 0⤊ 0⤋
The one opposite the longest side(9)
2007-05-12 13:36:38 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Use Law of Cosines
c2=a2+b2-2abCOS(C)
81=64+36-2(6)(8)COS(C)
81-100 = -96COS(C)
19 / 96 = COS(C)
ACOS(.1979) = C
C=78.7 degrees
2007-05-12 13:38:30 · answer #6 · answered by shanusav 2 · 0⤊ 0⤋