Answer:
∠J = 60°
Step-by-step explanation:
The Law of Cosines tells you ...
j² = k² +l² -2kl·cos(J)
Solving for J gives ...
J = arccos((k² +l² -j²)/(2kl))
J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)
J = 60°
_____
<em>Additional comment</em>
It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.
Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
or
[ ,
,
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ ,
,
]
In conclusion, the two unit vectors are;
[ ,
,
]
and
[ ,
,
]
<em>Hope this helps!</em>