HELP WITH THESE QUESTIONS PLEASEEE I WILL GIVE BRAINLIST AND 100 POINTS IVE TRIED VERY HARD AND CANT GET THE ANWSES NEED THESE DONE ASAP THANK YOU!
1.In triangle JKL, tan(b°) = three fourths and sin(b°) = three fifths. If triangle JKL is dilated by a scale factor of 3, what is cos(b°)?

triangle KL in which angle K is a right angle and angle L measures b degrees

cos(b°) = three fifths
cos(b°) = four fifths
cos(b°) = five thirds
cos(b°) = five fourths

2.If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z?

z = 3
z = 4
z = 5
z = 6

3.
Triangle BAC was dilated from triangle BDE at a scale factor of 2. What proportion proves that cos∠D = cos∠A?

Triangles BDE and BAC, in which angle B is a right angle, point D is between points B and A, and point E is between points B and C; BD measures 2 units, BE measures 3 units, and DE measures 3 and 61 hundredths units.

2 over 3 and 61 hundredths = 4 over 7 and 22 hundredths
three halves = six fourths
3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths
two thirds = four sixths

IMAGES ARE IN ORDER FIRST IMAGE IS FOR QUESTION 1 SECOND IMAGE IS FOR QUESTION 3 THERE IS NO IMAGE FOR QUESTION 2

Note that dilating a structure by whatever scale factor only affects the lengths of the sides of the Triangle, the angles of the Triangle remain the same.

Tan b remains = (3/4)

Sin b remains = (3/5)

And from trigonometric relations

(Sin b)/(cos b) = tan b

Cos b = (Sin b)/(tan b)

Cos b = (3/5) ÷ (3/4)

Cos b = (3/5) × (4/3) = (4/5)

Cos b = four fifths.

2) sin(9z − 1) = cos(6z + 1) in the range 0 < z ≤ 90

Note that cosine and some are related through cos θ = sin (90 - θ)

So, cos (6z + 1) = Sin [90 - (6z + 1)] = Sin (90 - 6z - 1)

sin(9z − 1) = cos(6z + 1)

Sin (9z - 1) = Sin (90 - 6z - 1)

We can then equate the angles

9z - 1 = 90 - 6z - 1

9z + 6z = 90°

15z = 90°

z = 6°

3) Triangle BDE is dilated by a scale factor of 2 to obtain triangle BAC.

Dilation by a scale factor of 2 means that all the sides of triangle BAC are twice as much as the corresponding sides of triangle BDE.

And triangle BDE is similar to triangle BAC.

It also means that the corresponding angles are necessarily equal.

BD = 2

BE = 3

ED = 3.61

BA = 2 × BD = 2 × 2 = 4

BC = 2 × BE = 2 × 3 = 6

CA = 2 × ED = 2 × 3.61 = 7.22

Cos ∠D according to trigonometric relations is given as (adj/hyp)

Adj = Adjacent side = BE = 3

Hyp = hypotenuse side = ED = 3.61

Cos ∠D = (3/3.61)

Cos ∠A can also be similarly obtained from trigonometric relations as (adj/hyp)

Adj = Adjacent side = BC = 6

Hyp = hypotenuse side = CA = 7.22

Cos ∠A = (6/7.22)

Since the two angles are corresponding angles of two similar triangles,

We can easily see that

Cos ∠D = Cos ∠A

(3/3.61) = (6/7.22) = 0.8310

Which is necessarily equal to each other

Hence, the proportion that proves that Cos ∠D = Cos ∠A is 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths.

Hope this Helps!!!

psm22415 psm22415 · Answer 2 · 2021-12-07T08:17:14+01:00

The value of Cosb° is four fifths.

The value of z is 6 degree.

The proportion that proves that Cos ∠D = Cos ∠A is 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths.

We have to determined

The values given in the following statements.

According to the question,

In triangle JKL,

Tanb° = three fourths = [tex]\frac{3}{4}[/tex]

Sinb° = three fifths = [tex]\frac{3}{5}[/tex]

Triangle JKL is then dilated by a scale factor of 3. The value of cosb° is,

The dilating a structure by whatever scale factor only affects the lengths of the sides of the Triangle,

The angles of the Triangle remain the same.

Tanb remains = [tex]\frac{3}{4}[/tex]

Sinb remains = [tex]\frac{3}{5}[/tex]

And from trigonometric relations cosb° is,

[tex]= \dfrac{sinb}{cosb} = Tanb\\\\= \dfrac{sinb}{tanb} = cosb\\\\cosb = \dfrac{\frac{3}{5}}{\frac{3}{4}}\\\\\\cosb = \dfrac{4}{5}[/tex]

The value of Cosb° is four fifths.

sin(9z − 1) = cos(6z + 1) in the range 0 < z ≤ 90

Cosine and some are related through cos θ = sin (90 - θ)

So, cos (6z + 1) = Sin [90 - (6z + 1)] = Sin (90 - 6z - 1)

sin(9z − 1) = cos(6z + 1)

Sin (9z - 1) = Sin (90 - 6z - 1)

Then, equate the angles

9z - 1 = 90 - 6z - 1

9z + 6z = 90°

15z = 90°

z = 6°

The value of z is 6 degree.

Triangle BDE is dilated by a scale factor of 2 to obtain triangle BAC.

Dilation by a scale factor of 2 means that all the sides of triangle BAC are twice as much as the corresponding sides of triangle BDE.

And triangle BDE is similar to triangle BAC.

The corresponding angles are necessarily equal.

BD = 2

BE = 3

ED = 3.61

Then,

BA = 2 × BD = 2 × 2 = 4

BC = 2 × BE = 2 × 3 = 6

CA = 2 × ED = 2 × 3.61 = 7.22

Cos ∠D according to trigonometric relations is given as,

Adjacent side = BE = 3

hypotenuse side = ED = 3.61

Then,

[tex]CosD = \dfrac{3}{3.61}[/tex]

Cos ∠A can also be similarly obtained from trigonometric relations as,

Adjacent side = BC = 6

hypotenuse side = CA = 7.22

Then,

[tex]CosA = \dfrac{6}{7.22}[/tex]

The two angles are corresponding angles of two similar triangles,

[tex]Cos D = Cos A\\\\\dfrac{3}{3.61} = \dfrac{6}{7.22} = 0.8310[/tex]

Which is necessarily equal to each other

Hence, the proportion that proves that Cos ∠D = Cos ∠A is 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths.

To know more about Trigonometry click the link given below.

https://brainly.com/question/16691385