A right circular cylinder has a volume of 500 cu in. If the base has a radius of 4 in., what's the altitude of the cylinder? Rou
1 answer:
You might be interested in
Answer:
m∠JKP = 31.5°
Step-by-step explanation:
Incenter of a triangle is the point where all the bisectors of interior angles intersect each other.
JN is the angle bisector of ∠KJL.
Therefore, m∠KJN = m∠LJN
(7x - 6) = (5x + 4)
7x - 5x = 6 + 4
2x = 10
x = 5
m∠KJN = (7x - 6)
= 7(5) - 6
= 35 - 6
= 29°
In ΔKJN,
m∠JKN + m∠KNJ + m∠NJK = 180°
m∠JKN + 90° + 29° = 180°
m∠JKN = 180°- 119° = 61°
Since KO is the angle bisector of ∠JKN,
m∠JKP =
=
= 30.5°