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
Answer:
There would be 4 stacks
Step-by-step explanation:
Heres how I did it.
4+4=8
8+4=12
12+4=16
Just keep adding 4 to it.
B 64/7 I did the math it is b
Answer:
4 and 5
Step-by-step explanation:
Assuming your 2nd sentence should be: "The sum of the squares of the two numbers is 41"
X + Y = 9 {equation 1}
X² + Y² = 41 {equation 2}
from equation 1: y = 9-x
Substitute that into equation 2
x² + (9-x)² = 41
x² + 81 - 18x + x²= 41
2x² - 18x + 40 = 0
We can factor out a 2
x² - 9x + 20 = 0
(x-5)(x-4) = 0
x = 5 or 4
checking
Your two numbers are 4 and 5
4² + 5²
= 16 + 25
= 41
