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Alenkasestr [34]
3 years ago
8

What is the simplest form of the expression below? cot(theta) cos(theta)/sin(theta)*tan(theta) divided by sin(theta)/cos(theta)

tan(theta)
Mathematics
2 answers:
-BARSIC- [3]3 years ago
7 0

Answer:

Step-by-step explanation:

B. cot(theta)

jeka57 [31]3 years ago
5 0
Cot(theta) is the one that works best for this problem
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1,290=h/10+h/5 What is h?​
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olga55 [171]

Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

3 0
3 years ago
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