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3 years ago

Find the exact value of each expression. sin75° cos(-75°)

Mathematics

2 answers:

suter [353]3 years ago

7 0

Answer:

Step-by-step explanation:

sin 75°

sin(30°+45°)

sin30°*cos45°+cos30°sin45°

1/2*1/ $\sqrt{2}$ + $\sqrt{3}$ /2*1 $\sqrt{2$

1/2 $\sqrt{2}$ + $\sqrt{3}$ /2 $\sqrt{2}$

1+3/2 $\sqrt{2}$

cos(-75°)

-cos(180°-105°)

-cos(-105°)

cos 105°

cos(45°+60°)

cos45°cos60°-sin45°sin60°

1/ $\sqrt{2}$ *1/2-1/ $\sqrt{2}$ * $\sqrt{3}$ /2

1/2 $\sqrt{2$ - $\sqrt{3$ /2 $\sqrt{2}$

1- $\sqrt{3}$ /2 $\sqrt{2}$

tino4ka555 [31]3 years ago

5 0

sin(75)

=0.96

cos(-75)

=0.25

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Find the angle between u = (-2,-5) and v = (5,2)

OleMash [197]

The angle between u = (-2,-5) and v = (5,2) is 134 degrees approximately.

<u>Solution:</u>

Given, two vectors are u = (-2, -5) and v = (5, 2)

We have to find the angle between two vectors.

We know that,

$a. b=\|a\| .\|b\| \times \cos \theta$

where $\theta$ is angle between vectors a and b

$\text { Now vectors are }(-2 \vec{\imath}-5 \vec{\jmath}) \text { and }(5 \vec{\imath}+2 \vec{\jmath})$

$(-2 \vec{\imath}-5 \vec{\jmath}) \cdot(5 \vec{\imath}+2 \vec{\jmath})=\sqrt{(-2)^{2}+(-5)^{2}} \times \sqrt{5^{2}+2^{2}} \times \cos \theta$

$\text { since }\|a\|=\sqrt{x^{2}+y^{2}} \text { for } a=x \vec{\imath}+y \vec{\jmath}$

$\begin{array}{l}{-10-10=\sqrt{29} \times \sqrt{29} \times \cos \theta} \\\\ {-20=29 \times \cos \theta} \\\\ {\cos \theta=\frac{-20}{29}} \\\\ {\theta=\cos ^{-1} \frac{-20}{29}} \\\\ {\theta=133.60}\end{array}$

Hence, the angle between given two vectors is 134 degrees approximately.

3 0

3 years ago

A 4-foot tall child walks directly away from a 12-foot tall lamppost at 2 mph. How quickly is the length of her shadow increasin

xz_007 [3.2K]

Answer:

The length of the shadow is increasing with the rate of 1.5 feet per sec

Step-by-step explanation:

Let AB and CD represents the height of the lamppost and child respectively ( shown below )

Also, let E be a point represents the position of child.

In triangles ABE and CDE,

$\angle ABE\cong \angle CDE$ ( right angles )

$\angle AEB\cong \angle CED$ ( common angles )

By AA similarity postulate,

$\triangle ABE\sim \triangle CDE$

∵ Corresponding sides of similar triangles are in same proportion,

$\implies \frac{AB}{CD}=\frac{BE}{DE}$

We have, AB = 12 ft, CD = 4 ft, BE = BD + DE = 6 + DE,

$\implies \frac{12}{4}=\frac{6+DE}{DE}$

$12DE = 24 + 4DE$

$8DE = 24$

$DE=3$

Now, the speed of walking = 2 mph = $\frac{2\times 5280}{3600}\approx 2.933\text{ ft per sec}$

Note: 1 mile = 5280 ft, 1 hour = 3600 sec

Thus, the time taken by child to reach at E

$= \frac{\text{Walked distance}}{\text{Walking speed}}$

$=\frac{6}{2.933}$

= 2.045 hours

Hence, the change rate in the length of shadow

$= \frac{\text{Length of shadow}}{\text{Time taken}}$

$=\frac{3}{2.045}$

= 1.5 ft per sec.

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3 years ago

I need answer quickly

vagabundo [1.1K]

Answer:

4.47

Step-by-step explanation:

4^2+b^2=6^2

16+b^2=36

-16. -16

√b^2=√20

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2 years ago

Which equation represents a line which is parallel to the line 2x+y=-6?

joja [24]

The answer is: [C]: " y = -2x - 1 " .
____________________________________________________

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3 years ago

What is the greatest number of cups that can be completely filled with water from the container

Elan Coil [88]

Answer:

Step-by-step explanation:

I have mathswatch work too do as well so its kinda annoying lol

so first we need to find the volume of the cuboid

36cm x 6cm x 23cm= 4968cm³

2/3x4968=3312cm³

3312/275=12.04

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3 years ago