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In triangle JKL, sin(b°) = 3/5 and cos(b°) =4/5. If triangle JKL is dilated by a scale factor of 2, what is tan(b°)?

a.) tan(b°) = 5/3
b.) tan(b°) = 4/3
c.) tan(b°) = 3/4
d.) tan(b°) = 5/4

In triangle JKL sinb 35 and cosb 45 If triangle JKL is dilated by a scale factor of 2 what is tanb a tanb 53 b tanb 43 c tanb 34 d tanb 54 class=

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xero099

Answer:

C. [tex]\tan (b^{\circ}) = \frac{3}{4}[/tex]

Step-by-step explanation:

For right triangles, trigonometric ratios are independent of scale factor, since internal angles do not change. That is:

[tex]\tan (b^{\circ}) = \frac{\sin b^{\circ}}{\cos b^{\circ}}[/tex] (1)

If we know that [tex]\sin (b^{\circ}) = \frac{3}{5}[/tex] and [tex]\cos (b^{\circ}) = \frac{4}{5}[/tex], then the tangent of the angle b is:

[tex]\tan (b^{\circ}) = \frac{3}{4}[/tex]

Correct answer is C.

The value of tan(b°) is given by the ratio of sin(b°) to cos(b°)

The correct response;

  • c.) tan(b°) = 3/4

Method by which the above response is obtained;

Given;

Given trigonometric ratios are;

[tex]sin(b^{\circ}) = \mathbf{\dfrac{3}{5}}[/tex]

[tex]\left \{ {{y=2} \atop {x=2}} \right. cos(b^{\circ}) = \mathbf{\dfrac{4}{5}}[/tex]

The given transformation = A dilation by a scale factor of 2

Required:

The value of tan(b°).

Solution:

The ratio of the sides following a dilation transformation is presented as follows;

  • [tex]\dfrac{\overline{JL}}{\overline{KL}} = \dfrac{2}{2} \times \dfrac{\overline{JL}}{\overline{KL}} = \dfrac{2 \cdot \overline{JL} }{2 \cdot\overline{KL}} = \mathbf{\dfrac{\overline{J'K'}}{\overline{K'L'}}}[/tex]

Therefore, the ratio of the sides corresponding sides in the preimage and

the image remain equal following a dilation transformation.

Which gives;

[tex]tan (b^{\circ})} = \dfrac{sin(b^{\circ})}{cos(b^{\circ})}} = \dfrac{\frac{3}{5} }{\frac{4}{5} } = \mathbf{\dfrac{3}{4}}[/tex]

The correct option is therefore;

  • c.) tan(b°) = 3/4

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