Question

In triangle JKL, tan(b°) = 3/4 and cos(b°) =4/5. If triangle JKL is dilated by a scale factor of 1/2, what is sin(b°)?

Answer 1

Answer:

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

Step-by-step explanation:

It is given that,

$\tan (b^\circ)=\dfrac{3}{4}$

$\cos (b^\circ)=\dfrac{4}{5}$

If a figure is dilated, then the image is similar to the figure. It means the corresponding angles of figure and image are congruent.  

So, the value of sin(b°) after dilation is equal to the value of sin(b°) before dilation.

We know that,

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

$\dfrac{\sin (b^\circ)}{\cos (b^\circ)}=\tan (b^\circ)$

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

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

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

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