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2 years ago

1.1 solve for x, where 0°< x 90°. write your answer to one decimal place. 1.1.1 tanx=sin38° 1.1.2cosec( x+10°)=1.345

Mathematics

1 answer:

snow_lady [41]2 years ago

5 0

The value of x in tan(x)=sin38° is 31.6 and the value of x in cosec(x+10°)=1.345 is 38.0

<h3>How to solve the trigonometry ratios?</h3>

The equations are given as:

tan(x)=sin38°

cosec( x+10°)=1.345

In tan(x)=sin38°, we have:

tan(x)=0.6157

Take the arc tan of both sides

x = 31.6

Also, we have:

cosec(x+10°)=1.345

Take the inverse of both sides

sin(x+10°) = 0.7434

Take the arc sin of both sides

x+10 = 48.0

Subtract 10 from both sides

x = 38.0

Hence, the value of x in tan(x)=sin38° is 31.6 and the value of x in cosec(x+10°)=1.345 is 38.0

Read more about trigonometry ratios at:

brainly.com/question/11967894

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Trigonometry tells you ...
tangent = opposite / adjacent
so you can write two relations involving LJ.
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The second equation can be used to write an expression for LJ that can be substituted into the first equation.
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tan(51°) = 3*tan(68°)/JM
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3 years ago

If sin theta equals 12 divided by 13 and theta is in quadrant 2 cos20 equals? and cos theta equals?

Anni [7]

Answers:

cos(2theta) = -119/169
cos(theta) = -5/13

=====================================================

The given info is:

sin(theta) = 12/13
theta is in quadrant II

From that, we can use the pythagorean trig identity to find that cos(theta) = -5/13. Keep in mind that cosine is negative in quadrant II.

Now use the trig identity below to compute cos(2theta)

$\cos(2\theta) = \cos^2(\theta)-\sin^2(\theta)\\\\\cos(2\theta) = \left(\frac{-5}{13}\right)^2-\left(\frac{12}{13}\right)^2\\\\\cos(2\theta) = \frac{25}{169}-\frac{144}{169}\\\\\cos(2\theta) = \frac{25-144}{169}\\\\\cos(2\theta) = -\frac{119}{169}\\\\$

Other options you could use are these identities

$\cos(2\theta) = 2\cos^2(\theta)-1\\\\$

$\cos(2\theta) = 1-2\sin^2(\theta)\\\\$

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3 years ago

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Arturiano [62]

Answer:

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Step-by-step explanation:

f(x)=-x^2+19

Let x= 3

f(x)=- (3)^2+19

= -9 + 19

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3 years ago

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