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solong [7]
3 years ago
7

(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 on

e 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

Mathematics
1 answer:
REY [17]3 years ago
5 0

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}

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