In ΔJKL, \overline{JL} JL is extended through point L to point M, m∠JKL = (2x+14)^{\circ}(2x+14) ∘ , m∠LJK = (3x+15)^{\circ}(3x+
15) ∘ , and m∠KLM = (7x+9)^{\circ}(7x+9) ∘ . What is the value of x?
1 answer:
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Answer:
x = 20, or 125
Step-by-step explanation:
We cannot tell the angle x is intended to refer to. So, here are two answers. Choose the one you think applies.
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The triangle that includes both marked angles is a right triangle. (The tangent meets the circle at right angles to the radius.) So, the angle x° at the external point is the complement of the marked central angle.
x° = 90° -70°
x = 20
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If x° is referring to the exterior angle associated with the isosceles triangle within the circle, then its measure is the supplement of the interior base angle of that triangle. It is ...
x° = 180° -(180° -70°)/2 = 90° +70°/2
x = 125
