Right triangles 1, 2, and 3 are given with all their angle measures and approximate side lengths. Use one of the triangles to approximate the ratio (KL)/(JL).

- 0.64
- 0.77
- 0.83
- 1.2

Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use
Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use

Answers

Answer 1

Answer:

[tex]\frac{KL}{JL}[/tex] = [tex]\frac{6.4}{7.7}[/tex] = 0.83

Step-by-step explanation:

The key understanding here is that ΔJKL is similar to triangle 3 based on the AA criterion (they both have a right angle and a 40° angle).

We can find [tex]\frac{KL}{JL}[/tex] by setting up a proportion statement that includes KL, JL, and the lengths of their corresponding sides in triangle 3.

We can use this proportion:

[tex]\frac{KL}{6.4}[/tex] = [tex]\frac{JL}{7.7}[/tex]

[tex]\frac{KL}{6.4}[/tex] ⇒ opposite to 40° angle

KL, JL ⇒ ΔJKL

[tex]\frac{JL}{7.7}[/tex] ⇒ adjacent to 40° angle

6.4, 7.7 ⇒ triangle 3

[Now see the attachment]

Now we can rewrite the equation to show the ratios of the side lengths within each triangle.

[tex]\frac{KL}{JL}[/tex] = [tex]\frac{6.4}{7.7}[/tex]

[tex]\frac{KL}{JL}[/tex] ⇒ ΔJKL

KL, 6.4 ⇒ adjacent to 40° angle

[tex]\frac{6.4}{7.7}[/tex] ⇒ triangle 3

JL, 7.7 ⇒ adjacent to 40° angle

Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use
Answer 2

Answer: it’s 0.82

Step-by-step explanation:


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[tex] \large \bigstar \frak{ } \large\underline{\sf{Solution-}}[/tex]

Consider, LHS

[tex]\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

We know,

[tex]\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} 

\\ \\ \text{So, using this identity, we get}

\\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

We know,

[tex]\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\ [/tex]

So, using this identity, we get

[tex]\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

can be rewritten as

[tex]\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}[/tex]

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[tex]\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}[/tex]

[tex]\rule{190pt}{2pt}[/tex]

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Answers

The transactions that would leave the initial balance the same are a deposit of $278 followed by a withdrawal of $278.

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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