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3 years ago

12

A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B an

d C, as shown below
What is the distance between the walls?
Answer
50 m
70 m
100 m
110 m

2 answers:

Gekata [30.6K]3 years ago

3 0

Answer:

Your answer would be B. 70 m good sir

Step-by-step explanation:

Vlada [557]3 years ago

3 0

Answer:

70

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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10,000/10=1,000 so the answer is C. 1/10.

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4 years ago

prove the following identity: sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x please provide a proof in some shape form or fash

wolverine [178]

Answer:

Step-by-step explanation:

Hello,

<u><em>Is this equality true ?</em></u>

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

<u>1. let 's estimate the left part of the equation</u>

$sec(x)csc(x)(tan(x) + cot(x)) =\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin(x)}{cos(x)}+\dfrac{cos(x)}{sin(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin^2(x)+cos^2(x)}{sin(x)cos(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{1}{sin(x)cos(x)})\\\\\\=\dfrac{1}{cos^2(x)sin^2(x)}$

<u>1. let 's estimate the right part of the equation</u>

<u /> $2+tan^2(x) + cot^2(x)=2+\dfrac{sin^2(x)}{cos^2(x)}+\dfrac{cos^2(x)}{sin^2(x)}\\\\=\dfrac{2cos^2(x)sin^2(x)+cos^4(x)+sin^4(x)}{cos^2(x)sin^2(x)}\\\\=\dfrac{(cos^2(x)+sin^2(x))^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1}{cos^2(x)sin^2(x)}$ <u />

This is the same expression

So

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

hope this helps

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