1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :
i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.
ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:
a^{2}=b ^{2}+c ^{2}-2bc(cosA)
2.
20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA) 400=81+169-234(cosA) 150=-234(cosA) cosA=150/-234= -0.641
3. m(A) = Arccos(-0.641)≈130°,
4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
8c + 6-3c -2
8c -3c + 6-2 = 5c + 4
Brainliest please!
Step-by-step explanation:
Pathagory and Theorem
a^2+b^2=c^2
(x-3)^2+(x-4)^2=6^2
expand:
2x^2-14x+25=36
2x^2-14x-11=0
x=(\sqrt{71}+7)/2
perimeter=(x-3)+(x-4)+6=2x-1
insert the value for x into 2x-1
+6=perimeter
Hope that helps :)
Answer:
1/8 cm = 0.125 cm = 1.25 mm
6 * (1.25 mm)^2 = 9.375 mm^2