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Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side

maria [59]

Answer:

cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = $-\frac{1}{3}$

cos(b) = $-\frac{1}{4}$

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = $-\sqrt{1-(-\frac{1}{3})^2}$ [Since, sin(a) = $\sqrt{(1-\text{cos}^2a)}$ ]

= $-\sqrt{\frac{8}{9}}$

= $-\frac{2\sqrt{2}}{3}$

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.

sin(b) = $-\sqrt{1-(-\frac{1}{4})^2}$

= $-\sqrt{\frac{15}{16}}$

= $-\frac{\sqrt{15}}{4}$

By substituting these values in the identity,

cos(a + b) = $(-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})$

= $\frac{1}{12}-\frac{\sqrt{120}}{12}$

= $\frac{1}{12}(1-\sqrt{120})$

= $\frac{1}{12}(1-2\sqrt{30})$

Therefore, cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

5 0

3 years ago

A car run 3/4 mile per minute another car can run 4/5 mile per minute.what is the difference in speed in miles per hour?

dybincka [34]

Answer:

3 miles per hour

Step-by-step explanation:

We need to subtract the two numbers to find the difference in speeds

4/5 - 3/4

We need to get a common denominator of 20

4/5 *4/4 - 3/4 *5/5

16/20 - 15/20

1/20

The car going 4/5 miles per minute is going 1/20 miles per minute faster than the car going 3/4 miles per minute.

The question asks for miles per hour, not miles per minute. We know that there are 60 minutes per hour.

1/20 miles per minute * 60 minutes / hour

60/20 miles/hour

3 miles/ hour

The car going 4/5 miles per minute is going 3 miles per hour faster than the car going 3/4 miles per minute.

3 0

3 years ago

Drag equations to each row to show why the equation of a straight line is y = 2x + 5, where 2 is the slope and 5 is the y-interc

Vedmedyk [2.9K]

Answer: y-5/x =2/1 for the two triangles on the graph are similar

Y-5=2x for multiple both sides by x

Y=2x+5 for add 5 to both sides

Step-by-step explanation:

3 0

3 years ago

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

1 year ago

Naomi has 50 red beads and white beads. The number of red beads is 1 more than 6 times the number of white beads. How Many red b

Tanya [424]

This is confusing for me

5 0

3 years ago