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3 years ago

13

....Math I need help

1 answer:

Brilliant_brown [7]3 years ago

3 0

M<FIB = 90 - (4x+3)
m<FIB = 87 - 4x

---------------------
m<HFG = 90 - (5x + 4)
m<HFG = 86 - 5x

-------------------------------
<BCF = 180 - (<CBF + <CFB)
<CBF = 10x + 3
<CFB =90 - <CFD
but <CFD = <HFG = 86 - 5x (vertical angles)

<CFB =90 -(86 - 5x) = 4 + 5x

<BCF = 180 - (10x + 3 + 4 + 5x)
<BCF = 180 - (15x + 7)
<BCF = = 173 - 15x

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3 years ago

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3 years ago

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Step-by-step explanation:

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3 years ago

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denis-greek [22]

Answer:

<u>______________________________________________________</u>

<u>TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-</u>

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

<u>SOME GENERAL TRIGNOMETRIC FORMULAS :-</u>

<u></u> $\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

<u></u> $\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

<u>______________________________________________________</u>

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

3 0

3 years ago