In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is tan(b°)? triangle KL in which angle K is a right angle and angle L measures b degrees tan(b°) = five thirds tan(b°) = four thirds tan(b°) = three fourths tan(b°) = five fourths

Tan(b°) = 3/4 = three fourths (C)

Step-by-step explanation:

Triangle JKL is a right angled triangle in which angle K is a right angle and angle L= b°

We would apply SOHCAHTOA from trigonometry to determine the value for each sides.

For ∆JKL

Sin(b°) = opposite/hypotenuse

Sin(b°) = 3/5

Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4

From the above we know the value of each sides of ∆JKL

Find attached the diagram of the above triangle (1)

∆JKL is dilated by a scale factor = 2

Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:

Opposite = 2(3) = 6

Adjacent = 2(4) = 8

Hypotenuse = 2(5) = 10

To find tan(b°) of the new triangle, we would apply tangent ratio

Tan(b°) = opposite/adjacent

Tan(b°) = 6/8

Tan(b°) = 3/4

Find attached the diagram for the new triangle (2)

Diagram 3 shows the drawing of both triangles together.

From the above, we can see the angle doesn't change when a shape is dilated by a scale factor.

Therefore, tan(b°) = three fourths (C)

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225