Question

(05.01 MC) 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°)? PLEASE HELP! 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

Answer:

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

Step-by-step explanation:

We are given a $\triangle JKL$ where $\angle K = 90^\circ$

$tan(b^\circ) = \dfrac{3}{4}\\cos(b^\circ) = \dfrac{4}{5}$

As per trigonometric rules,

$tan(\theta) = \dfrac{Perpendicular}{Base}$

$cos(\theta) = \dfrac{Base}{Hypotenuse}$

Comparing the given value of tangent and cosine:

Perpendicular = 3 units

Base = 4 units

Hypotenuse = 5 units

Also, we know that:

$sin(\theta) = \dfrac{Perpendicular}{Hypotenuse}$

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

Now, it is given that the triangle is dilated by a scale factor of $\frac{1}{2}$.

i.e. all the sides will become half of its initial values.

New Perpendicular = 1.5 units

New Base = 2 units

New Hypotenuse = 2.5 units

Using the formula for sine:

$sin(\theta) = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin(b^\circ)= \dfrac{1.5}{2.5}\\\Rightarrow sin(b^\circ)= \dfrac{15}{25}\\\Rightarrow sin(b^\circ)=\dfrac{3}{5}$