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ludmilkaskok [199]
2 years ago
7

Help please...........geometry

Mathematics
1 answer:
r-ruslan [8.4K]2 years ago
5 0
I am going with major arch ABC
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Which basic trigonometric identity would you use to verify that tan x cos x =sin x
True [87]

Answer:

\huge\boxed{\tan x=\dfrac{\sin x}{\cos x}}

Step-by-step explanation:

\tan x\cos x=\sin x\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\text{Therefore}\\\\\tan x\cos x=\dfrac{\sin x}{\cos x}\cdot\cos x\qquad\text{cancel}\ \cos x\\\\\tan x\cos x=\sin x\\\\\text{For}\ x\neq\dfrac{\pi}{2}+k\pi},\ k\in\mathbb{Z}

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3 years ago
Two diameters of circle P are AB and CD. Find the measure of arc ACD and arc AC when arc AD has a measure of 20 degrees
steposvetlana [31]

Answer:

Hence measure of arc ACD and arc AC are 340 degrees and 160 degrees respectively.

Step-by-step explanation:

Given:

A Circle with p as center and AB and CD as 2 diameters for a circle.

To Find:

Measure of ACD, and arc AC when arc AD is 20 degrees.

Solution:

<em>As we know that , Diameter is biggest chord which passes through center of circle,</em>

<em>Hence it divides the circle into two equal parts and making center as midpoint of it</em> .

(Refer the attachment)

angle APD=20 degree or arc AD=20 degree.

So AB as diameter ,so  arc DB=160 degree  or BPD=160 degree as A-B-P are

and angle APD and BPC are vertical opposite angles so they are congruent BPC=20 degree

Now,

points A,C,B,D,A makes circle which measures 360 degrees.

to find ACD

measure of ACD=360-(meaure of arc AD or angle APD)

=360-20

=340 degree.

Now arc AC,

as AB is diameter with p as midpoint ,A-P-B

arc APB=180 and arc BC=20 or angle BPC=20 degrees.

For measure of arc AC= measure of arc APB- arc BC

arc AC=180-20 degrees.

arc AC=160 degrees.

5 0
3 years ago
Please I need help!!!
AnnyKZ [126]

Answer:

22.5

Step-by-step explanation:

Because

180÷2=90

so that would mean

45÷2=22.5

5 0
2 years ago
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Hannah put $2,000 in a savings
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He measured the height of the box to be 7 in. Then, Stephan drew a line from the center of one of the hexagons to each of its ve
JulijaS [17]
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