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AveGali [126]
3 years ago
12

....................answer

Mathematics
2 answers:
Oksana_A [137]3 years ago
8 0

Answer:

d. 2

Step-by-step explanation:

Ludmilka [50]3 years ago
5 0
D) 2

Explanation : (2•2+3•4)=16
16/4=4
4-2=2
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GREYUIT [131]

Question 7: Option 1:  x = 33.5°

Question 8: Option 3: x = 14.0°

Step-by-step explanation:

<u>Question 7:</u>

In the given figure, the value of perpendicular and hypotenuse is given, so we have to use any trigonometric ratio to find the value of angle as the given triangle is a right-angled triangle

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Perpendicular = P = 32

Hypotenuse = H = 58

So,

sin\ x = \frac{P}{H}\\sin\ x = \frac{32}{58}\\sin\ x = 0.5517\\x = sin^{-1} ( 0.5517)\\x =33.48

Rounding off to nearest tenth

x = 33.5°

<u>Question 8:</u>

In the given figure, the value of Base and Perpendicular is given, we will use tangent trigonometric ratio to find the value of x

So,

Perpendicular = P = 5

Base = B = 20

So,

tan\ x = \frac{P}{B}\\tan\ x = \frac{5}{20}\\tan\ x = 0.25\\x = tan^{-1} (0.25)\\x = 14.036

Rounding off to nearest tenth

x = 14.0°

Keywords: Right-angled triangle, trigonometric ratios

Learn more about trigonometric ratios at:

  • brainly.com/question/909731
  • brainly.com/question/902892

#LearnwithBrainly

8 0
3 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
3 years ago
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