In ΔJKL, k = 22 cm, ∠L=23° and ∠J=26°. Find the length of j, to the nearest 10th of a centimeter.

We have been given that in ΔJKL, k = 22 cm, ∠L=23° and ∠J=26°. We are asked to find the length of j to the nearest tenth of a centimeter.

We will use law of sines to solve for side j.

, where, a, b and c are opposite sides to angles A, B and C respectively.

We need to find measure of angle K to apply law of sine to our given problem.

Using angle sum property, we will get:

Upon rounding to nearest tenth, we will get:

Therefore, the length of j is approximately 12.8 cm.

the answer is: the length is 5/6 yards.

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step-by-step explanation:

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.