# Triangle questions :S

• Apr 24th 2007, 03:08 PM
LoveDeathCab
Triangle questions :S
Hey, yet another question for you!

2. Triange ABC has AB=10cm, BC=7cm and angle B=80degree's Calculate
a) the area of the triangle
b) the length of CA
c) the size of the angle C

For a) i got 34.47cm^3
For b) i got 3.34cm
I can't figure out c)

If anyone can check my answers and help me with c) i'd be most grateful! :D
• Apr 24th 2007, 03:33 PM
Jhevon
Quote:

Originally Posted by LoveDeathCab
Hey, yet another question for you!

2. Triange ABC has AB=10cm, BC=7cm and angle B=80degree's Calculate
a) the area of the triangle

Use the formula,
Area = (1/2)acsinB
= (1/2)(10)(7)sin80
= 35sin80
= 34.49 cm^2
you were right, your units were wrong!:D

Quote:

b) the length of CA
Use the law of cosines:

(CA)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)cosB
=> (CA)^2 = 10^2 + 7^2 - 2(10)(7)cos80
=> (CA)^2 = 149 - 24.31074487
=> (CA)^2 = 124.6892551
=> CA = sqrt(124.6892551)
=> CA = 11.17 cm

seems you were incorrect:(
what did you do to get your answer for this one?

Quote:

c) the size of the angle C
Now that we know all the sides, we could use the cosine rule again, but for the sake of variety, let's use the sine rule a.k.a. the law of sines

Now b/sinB = c/sinC
=> AC/sinB = AB/sinC
=> sinC = AB/(AC/sinB)
=> C = arcsin[AB/(AC/sinB)]
=> C = arcsin[10/(11.17/0.984807753)]
=> C = arcsin(0.88165421)
=> C = 61.84 degrees
• Apr 24th 2007, 03:36 PM
LoveDeathCab
Thanks very much, i looked over my old workings on this question and it seems i got it right in the first place! :D