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Triangle J K L is shown. The length of L J is 15. Angle K L J is 28 degrees and angle L J K is 110 degrees. Determine the measures of all unknown angles and side lengths of △JKL. Round side lengths to the nearest hundredth. m∠K = ° JK ≈ LK ≈

Triangle J K L is shown The length of L J is 15 Angle K L J is 28 degrees and angle L J K is 110 degrees Determine the measures of all unknown angles and side class=

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Answer:

∠JKL = 42°

JK = 10.52

LK = 21.07

Step-by-step explanation:

We will use sine rule to find length of unknown sides.

The sin rule is:

[tex]\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}[/tex]

First, lets find the unknown angle ∠JKL.

We know sum of all 3 angles in a triangle is 180, so

28 + 110 + ∠JKL = 180

∠JKL = 180 - 28 - 110 = 42°

Now using sine rule, we find JK:

[tex]\frac{Sin28}{JK}=\frac{Sin42}{15}\\JKSin42=15Sin28\\JK=\frac{15Sin28}{Sin42}\\JK=10.52[/tex]

Now using sin rule, we find LK:

[tex]\frac{Sin42}{15}=\frac{Sin110}{LK}\\LKSin42=15Sin110\\LK=\frac{15Sin110}{Sin42}\\LK=21.07[/tex]

In ΔJKL: m∠K = 42°,  JK ≈ 10.52 units,  LK ≈ 22.08 units

Given:

∠LJK = 110°

∠KLJ = 28°

Side JL = 15 units

To find: ∠JKL = m∠K

We know that the sum of all the angles of a triangle is equal to 180°

∠LJK + ∠KLJ + ∠JKL = 180°

110° +  28° + ∠JKL = 180°

∠JKL = 180° - 138°

∠JKL = 42°

To find the side we use the Sine rule:

As per Sine rule:

[tex]\frac{a}{Sin A} =\frac{b}{Sin B}=\frac{c}{Sin C}[/tex]

Where a, b and c are the side opposite to ∠A, ∠B, and ∠C respectively.

Using the Sine rule in the given triangle

[tex]\frac{a}{sin 110} =\frac{15}{sin 42}=\frac{c}{sin 28}[/tex]

Solving:

[tex]\frac{a}{sin 110} =\frac{15}{sin 42}\\\\a=\frac{15}{sin 42}(sin 100)\\a=22.08 units[/tex]

Solving:

[tex]\frac{15}{sin 42}=\frac{c}{sin 28}\\\\c=\frac{15}{sin 42}(sin 28) \\c=10.52 units[/tex]

Therefore, m∠K = 42°,  JK ≈ 10.52 units  LK ≈ 22.08 units

For more information:

https://brainly.com/question/22288720

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