simple TRIG help . please!!!?

Question

really basic things again , i don't know how to do..

:(

Determine the measure of angle A in ABC to the nearest degree, if a =55cm, b=26cm and c=32 cm. (A=143*)

Find the area of triangle XYZ to the nearest tenth of a square metre, if x=9.5m, y=10.3, and z=6.8m. (Answer=31.3m^2)

Answer 1

To your second question:

Calculate it by Heron's formula. You may find it here:

http://www.btinternet.com/~se16/hgb/triangle.htm

Answer 2

for the first one, the easiest way to go is proportions.
i'll show you
if a triangle can only have 180* total, and the opposite side lengths are proportional to the angle then we can say that-
{55/(55+26+32)}=A/180
55/113=A/180
A=87.6 (about)
is the answer supposed to be 143*? cuz i dont think thats right

after going through a few ways of doing the second one, i think the best way is this-
Area = sqrt(s*(s-a)*(s-b)*(s-c))
where s=(a+b+c)/2

it ends up being an extremely long equation, but it simplifies nicely. if i were to type it, it'd take for ever!
hope i helped!!
-M

Answer 3

You would help yourself a great deal if you learned how to solve these problems. Try an internet search for 'trigonometry formulas', there are several sites out there like http://mathforum.org/dr.math/faq/formulas/faq.trig.html .

For (1), I would suggest the law of cosines:
a^2 = b^2 + c^2 - 2bc cos(A)

For (2), try
Area = sqrt( s (s-a)(s-b)(s-c))
where s = (a + b+ c)/2

Answer 4

for the question for the area of XYZ.

the area of a triangle is ½(B*H) {½ Base times height}
½(6.8 x 9.5) but the answer i got was 32.3m^2.

i hope this helps.