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3 years ago

Select the true statement about triangle ABC 13 A. sin A = cos C B. sin A = tan C C. sin A = sin C D. sin A = cos B

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:

sin C = cos A

Step-by-step explanation:

Just did it!

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Make w the subject of the relation p=2 (k+w)

ICE Princess25 [194]

Answer:

hope that will help you

3 0

3 years ago

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I NEED HELP!!<br>answer from the image below <br><br>m∠BCD =

nevsk [136]

Answer:

28 degrees

Step-by-step explanation:

If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.The corresponding angles are the ones at the same location at each intersection.

In the figure ABD and EDF are corresponding angles. So they are equal

So equation the angle ABD and EDF , we get

(3x+4) = (7x-20)

Group the like terms,

3x-7x = -20-4

-4x = -24

$x = \frac{-24}{-4}$

x = 6

Thus BCD will be,

(6x - 8)

=>(6(6)-8)

=>(36-8)

=> 28 degrees

4 0

3 years ago

What is the exact distance from (−4, −2) to (4, 6)?

Gekata [30.6K]

The distance between any two points is:

d^2=(x2-x1)^2+(y2-y1)^2

d^2=(6--2)^2+(4--4)^2

d^2=8^2+8^2

d^2=64+64

d^2=128

d=√128 units

6 0

4 years ago

Lena bought 14 yards of ribbon. How many feet of ribbon did she buy?

IgorLugansk [536]

$[Hello,BrainlyUser!\; I'm\;Lenvy]$

$Your\;Answer\;Is\;Below$

Answer:

42 feet

Step-by-step explanation:

First let know what is

Yards to Feet

Note that;

1 Yard ⇒ 3 Feet

Thus, Formula: Multiply the length value by 3

Given:

Lena bought 14 yards of ribbon. How many feet of ribbon did she buy?

Therefore, 14 * 3 = 42

Hence, Answer is Lena brough 42 feet of ribbon.

~[CloudBreeze]~

4 0

3 years ago

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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

4 years ago

Select the true statement about triangle ABC 13 A. sin A = cos C B. sin A = tan C C. sin A = sin C D. sin A = cos B​

Select the true statement about triangle ABC 13 A. sin A = cos C B. sin A = tan C C. sin A = sin C D. sin A = cos B