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:

2.

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:
x = 25
Step-by-step explanation:
We know that the remaining angles add up to 90°, because the right angle (triangles have 180° total.)
2x + 15 + x = 90 | Given
3x + 15 = 90 | Combine x
3x = 75 | Subtract 15
x = 25
44x10^-2
or in other words
44 times 10 to the -2nd power
We can solve this by finding one of the numbers first
Let the smaller number be y.
Since these 2 numbers are consecutive odd numbers, the larger number should be 2 more than the smaller one (y+2).
Therefore,
y + (y+2) = 284
2y +2 = 284
shift +2 to the other side and turn it into -2
2y = 284 - 2
2y = 282
shift x2 to the other side and turn it into /2
y = 282/2
y = 141
Now we got the smaller number, which is 141, we can also find the larger number by adding 2 to it. 141+2 = 143,
Therefore, your answer is 141 and 143