Question

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

Answer 1

The angle of sine has a value of 3/5.

Option A is the correct representation of angle sin(b°).

How do you calculate the angle b for sine?

Given that tan(b°) = 3/4 and cos(b°) = 4/5.

In trigonometry, we know that,

$\dfrac {sin \theta}{cos \theta} = tan \theta$

The angle of sine can be written as below.

$sin \theta = cos \theta \times tan \theta$

For the angle b, the above expression can be written as,

$sin (b^\circ) = cos (b^\circ) \times tan (b^\circ)$

Substituting the values in the above expression, we get,

$sin (b^\circ) = \dfrac {3}{4} \times \dfrac {4}{5}$

$sin (b^\circ) = \dfrac {3}{5}$

Hence we can conclude that the angle of sine has a value of 3/5.

To know more about the angle of sine, follow the link given below.

brainly.com/question/2512010.