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°) = three fourths

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its sides is proportional and its corresponding angles are congruent

Let

J'K'L' > the dilation of triangle JKL

Triangles J'K'L' and JKL are similar and their corresponding angles are congruent

Remember that

tan(b°)=sin(b°)/cos(b°)

substitute the given values

tan(b°)=(3/5)/(4/5)=3/4

therefore

tan(b°) = three fourths

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2