Could you guys show the workings
2007-10-14 05:59:32 · 6 answers · asked by Napster 2 in Science & Mathematics ➔ Mathematics
9/sin 40 = 5/sin C
C = arcsin (5sin (40)/9)= 22.797 degrees
So angle B = 180 - 40 - 22.797 = 117.203 degrees
AC = 9sin 117.203/sin40 = 12.453 cm
2007-10-14 06:18:48 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
The altitude to BC is drawn on the midpoint of BC it is 5 and its length is comparable to the ?(13^2 - 5^2) or the ?one hundred forty four it is 12. Angles B and C are equivalent to the sine^-a million (12/13) it is sixty seven.38º . the perspective BAC is 40 5.24º,The tan 40 5.24º (a million.008) x. 6.5 = the size of the perpendicular bisector of AB.That length is 6.fifty 5. in case you like the altitude from AB to indicate C, that line is comparable to the sin 40 5.24º circumstances 13 = 9.23
2016-10-22 09:23:01 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Use the law of cosine,
AC^2 = 5^2 + 9^2 -2(5)(9)cos(40)
AC = 6.087 cm
2007-10-14 06:17:28 · answer #3 · answered by sahsjing 7 · 0⤊ 1⤋
Use the law of sines. Although, I'm a little too lazy to actually do the problem out on the computer sorry. But you can do it, just plug the numbers into the equation: a/sinA = b/sinB = c/sinC
2007-10-14 06:07:42 · answer #4 · answered by Anonymous · 0⤊ 0⤋
this can be done using the SSA theorem, that's all I remember you should look that up maybe it'll help
2007-10-14 06:08:48 · answer #5 · answered by Anonymous · 0⤊ 0⤋
hi
you have to use the formula
cos(BAC)={(AB)^2+(AC)^2-(BC)^2}/2(AB)(AC)
2007-10-14 06:13:26 · answer #6 · answered by Anonymous · 0⤊ 0⤋