Mathematics: (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

(05.01 MC) In triangle JKL, tan(b°) = three fourths

kelly1027

01.10.2020

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

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:

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

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

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liddopiink1

29.10.2020

100 kilometers

Step-by-step explanation:

In her first week of training, she runs = 10 km

In her second week she runs = 15 km.

In her third week she runs = 25 km

In fourth week she runs = 50 km.

Total distance she run in her training = 10 + 15 + 25 + 50

= 100 km.

She runs 100 kilometers in total in her training before race.

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