I need help please!!!!!!!!!!!!!
2 answers:
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9514 1404 393
Answer:
not tangent
Step-by-step explanation:
The triangle has side lengths 5, 12, 14. You can compute the "form factor" for the triangle:
a^2 +b^2 -c^2 . . . . . . . . where 'c' is the longest side
= 5^2 +12^2 -14^2
= 25 +144 -196 = -27
The negative value tells you this triangle is obtuse. If AB were a tangent, it would be perpendicular to the radius—the triangle would be a right triangle.
If you compare this calculation to the Pythagorean theorem, you see that the length AB is longer than the length √(25+144) = 13 that is necessary for the triangle to be a right triangle. That means the angle at A is greater than 90°.
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<em>Additional comment</em>
This "form factor" calculation is part of the calculation you would do using the Law of Cosines to determine the largest angle. The sign of the "form factor" tells you the sign of the cosine of the angle. Angles whose cosine is negative are greater than 90°. A positive "form factor" indicates an acute triangle.
I find computing the "form factor" in this way makes interpretation of the result fairly easy. For me, it eliminates the confusion I had when the numbers of the Pythagorean theorem didn't add up. For me, it was too much work to figure whether the triangle was acute or obtuse, and I often got it wrong.
For those interested, the angle measure is arccos((a^2+b^2-c^2)/(2ab)). Here, that's arccos(-27/(2·5·12)) ≈ 103.003°.
Answer:
The vector equation of the line is and parametric equations for the line are
,
,
.
Step-by-step explanation:
It is given that the line passes through the point (0,11,-8) and parallel to the line
The parametric equation are defined as
Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.
From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.
The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are
Vector equation of a line is
where, is a position vector and
is cosine of parallel vector.
Therefore the vector equation of the line is and parametric equations for the line are
,
,
.