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3 years ago

Use ΔABC to find the value of cos B A) 12/13 B) 12/5 C) 5/13 D) 5/12

Mathematics

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

C) 5/13

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that the cosine ratio is ...

Cos = Adjacent/Hypotenuse

For the given problem, this means ...

cos(B) = BC/BA = 15/39

cos(B) = 5/13

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The area of a parrellogram is 31.5ft and height is 6.3ft. What is the length of the base

Musya8 [376]

Answer:

Area = base × height

31.5 = base × 6.3

base = 5 ft

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2 years ago

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Prove :<br>sin²θ + cos²θ = 1<br><br><br>thankyou ~

Gnom [1K]

Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

$\sf\longrightarrow \bf 1 = RHS$

Since LHS = RHS ,

Hence Proved !

I hope this helps.

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1 year ago

What is the range of h?

Bas_tet [7]

Answer:

-3 ≤y≤6

Step-by-step explanation:

The range is the output values or the y values

The y values go from -3 to 6 including these values

-3 ≤y≤6

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3 years ago

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Answer:

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2 years ago

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Combine the following expressions.

Keith_Richards [23]

Answer:

second option

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Simplifying the radicals

$\sqrt{27x^{3} }$

= $\sqrt{9(3)x^2.x}$

= $\sqrt{9}$ × $\sqrt{3}$ × $\sqrt{x^2}$ × $\sqrt{x}$

= 3 × $\sqrt{3}$ × x × $\sqrt{x}$

= 3x $\sqrt{3x}$

-------------------------------

$\sqrt{12x^{3} }$

= $\sqrt{4(3)x^2.x}$

= $\sqrt{4}$ × $\sqrt{3}$ × $\sqrt{x^2}$ × $\sqrt{x}$

= 2 × $\sqrt{3}$ × x × $\sqrt{x}$

= 2x $\sqrt{3x}$

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Thus

3 × 3x $\sqrt{3x}$ - 2 × 2x $\sqrt{3x}$

= 9x $\sqrt{3x}$ - 4x $\sqrt{3x}$

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2 years ago