=0 [/tex]
=0 [/tex]
(Reciprocal Identity)
(ZeroProduct Property)
and
x=0 and x=
are the solutions.
Check the picture below.
notice the sides, now, on the second triangle, side 6 slants a bit more to fit in 13, on the third triangle, side 6 slants even further to fit 13 in, now, if 6 were to slant completely, it'll make a flat-line with side 5, and there will be a triangle no more.
but even if side 6 would stretch to a flat-line, 5+6 is just 11, whilst side 13 is longer than that, so no dice.
Answer:
about 1.56637 radians ≈ 89.746°
Step-by-step explanation:
The reference angle in radians can be found by the formula ...
ref angle = min(mod(θ, π), π -mod(θ, π))
Equivalently, it is ...
ref angle = min(ceiling(θ/π) -θ/π, θ/π -floor(θ/π))×π
<h3>Application</h3>
When we divide 11 radians by π, the result is about 3.501409. The fractional part of this quotient is more than 1/2, so the reference angle will be ...
ref angle = (1 -0.501409)π radians ≈ 1.56637 radians ≈ 89.746°
__
<em>Additional comment</em>
For calculations such as this, you need to use the most accurate value of pi available. The approximations 22/7 or 3.14 are not sufficiently accurate to give good results.
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