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2 years ago

If anyone can help me to solve this.

Mathematics

2 answers:

MissTica2 years ago

4 0

Answer:

$\boxed {1) 90^{o}}\\\boxed {2) 80^{o}}$

Step-by-step explanation:

Part 1 :

x + 96 + x + 96 = 180 (angle pairs of a line)
2x + 192 = 180
2x = -12
x = -6

∠ = -6 + 96
∠ = 90°

Part 2 :

x + 109 + x + 89 = 180 (same reason)
2x + 198 = 180
2x = -18
x = -9

∠ = -9 + 89
∠ = 80°

Sindrei [870]2 years ago

3 0

Answer:

x + 96 = 90°

x + 89 = 80°

Step-by-step explanation:

Consecutive Interior Angles Theorem

When a transversal line intersects two parallel lines, it forms two pairs of consecutive angles on either side of the transversal line. Each pair of consecutive interior angles are supplementary (sum to 180°).

Question 1

To find the measure of the angle, find the value of x by using the Consecutive Interior Angle Theorem:

⇒ (x + 96)° + (x + 96)° = 180°

⇒ x + 96 + x + 96 = 180

⇒ 2x + 192 = 180

⇒ 2x + 192 - 192 = 180 - 192

⇒ 2x = -12

⇒ 2x ÷ 2 = -12 ÷ 2

⇒ x = -6

Substitute the found value of x into the expression to find the measure of the angle indicated in bold:

⇒ x + 96 = -6 + 96 = 90°

Question 2

To find the measure of the angle, find the value of x by using the Consecutive Interior Angle Theorem:

⇒ (x + 109)° + (x + 89)° = 180°

⇒ x + 109 + x + 89 = 180

⇒ 2x + 198 = 180

⇒ 2x + 198 - 198 = 180 - 198

⇒ 2x = -18

⇒ 2x ÷ 2 = -18 ÷ 2

⇒ x = -9

Substitute the found value of x into the expression to find the measure of the angle indicated in bold:

⇒ x + 89 = -9 + 89 = 80°

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3 years ago

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The angle,2Θ, lies in the third quadrant such that cos2Θ=-2/5. Determine an exact value for tanΘ . Show your work including any

horrorfan [7]

Answer:

$tan(\theta)=\frac{\sqrt{21}}{3}$

Step-by-step explanation:

1. Approach

One is given the following information:

$cos(2\theta)=-\frac{2}{5}$

One can rewrite this as:

$cos(2\theta)=-0.4$

Also note, the problem says that the angle ( $2\theta$ ) is found in the third quadrant.

Using the trigonometric identities ( $cos(2\theta)=2(cos^2(\theta))-1$ ) and ( $cos(2\theta)=1-2(sin^2(\theta))$ ) one can solve for the values of ( $cos(\theta)$ ) and ( $sin(\theta)$ ). After doing so one can use another trigonometric identity ( $tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$ ). Substitute the given information into the ratio and simplify.

2. Solve for $(cos(\theta))$

Use the following identity to solve for ( $cos(\theta)$ ) when given the value ( $cos(2\theta)$ ).

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

Substitute the given information in and solve for ( $cos(\theta)$ ).

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

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Inverse operations,

$-0.4=2(cos^2(\theta))-1$

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Since this angle is found in the third quadrant its value is actually:

$cos(\theta)=-\sqrt{0.3}$

3. Solve for $(sin(\theta))$

Use the other identity to solve for the value of ( $sin(\theta)$ ) when given the value of ( $cos(2\theta)$ ).

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

Substitute the given information in and solve for ( $sin(\theta)$ ).

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

$-0.4=1-2(sin^2(\theta))$

Inverse operations,

$-0.4=1-2(sin^2(\theta))$

$-1.4=-2(sin^2(\theta))$

$0.7=sin^2(\theta)$

$\sqrt{0.7}=sin(\theta)$

Since this angle is found in the third quadrant, its value is actually:

$sin(\theta)=-\sqrt{0.7}$

4. Solve for $(tan(\theta))$

One can use the following identity to solve for $(tan(\theta))$ ;

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Substitute the values on just solved for and simplify,

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$tan(\theta)=\frac{-\sqrt{0.7}}{-\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{0.7}}{\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

Rationalize the denominator,

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}*\frac{\sqrt{\frac{3}{`0}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}*\frac{3}{10}}}{\sqrt{\frac{3}{10}*\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{21}{100}}}{\frac{3}{10}}$

$tan(\theta)=\frac{\frac{\sqrt{21}}{10}}{\frac{3}{10}}$

$tan(\theta)=\frac{\sqrt{21}}{10}*\frac{10}{3}$

$tan(\theta)=\frac{\sqrt{21}}{3}$

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3 years ago

If anyone can help me to solve this.​

If anyone can help me to solve this.