Question

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Answer this question ===>>

In right angled ∆ ABC , angle B = 90 and ratio of BC to AC is 1 : 3.
then Find the value of..

$\frac{4 \tan(a) - 5 \cos(a) }{2 \cos(a) + 4 \cot(a) } \\ \\ \\ \\ answer \: = \: \frac{ - 1}{4}$
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Answer 1

HEY

Given that :-
/_ B =90°
Let the ratio constant be x .
So , BC = x
AC = 3x .
By pythagoras theorem ,
AB = √ ( 3x ) ² - ( x ) ²
AB = √ 9x² - x²
AB = √8x²
AB = 2x √2

Now ,
Tan( a ) = p / b = BC / AB
= x / 2x√2 = 1 / 2√2

Cos ( a ) = b / h = AB / AC
= 2x√2 / 3x = 2√2 / 3

Cot ( a ) = 1 / tan ( a )
= 2√2 .

Now ,

4tan ( a ) - 5 Cos ( a ) / 2Cos ( a) + 4 Cot(a)
Puting values , we get ,

4 * 1 / 2√2 - 5 * 2√2 / 3 divided by 2 * 2√2 / 3 + 4 * 2√2

= √ 2 - 10 √2 / 3 / 4√2 / 3 + 8√2
= 3√2 - 10√2 / 3 / 4√2 + 24√2 / 3
= -7√2 / 28√2
= -7 / 28
= - 1 / 4


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Answer 2

HELLO DEAR,

given that:-

<B=90°

BC/AC=1/3

=> BC=X AND AC=3X

IN∆ABC ,<B=90°

AC²=AB²+BC²

=> (3X)²=AB²+(X)²

=> AB²=9X²-X²=8X²

=> AB=2X√2

now,

tanA=1/cotA=1x/2x√2 = 1/2√2

AND

cosA=2x√2/3x = 2√2/3

$\frac{4 \tan(a) - 5 \cos(a) }{2 \cos(a) + 4 \cot(a) } \\ \\ = > \frac{4 \times \frac{1}{2 \sqrt{2} } - 5 \times \frac{2 \sqrt{2} }{3} }{2 \times \frac{2 \sqrt2 }{3} + 4 \times 2 \sqrt{2} } \\ \\ = > \frac{ \sqrt{2} - \frac{10 \sqrt{2} }{3} }{ \frac{4 \sqrt{2} }{3} + 8 \sqrt{2} } \\ \\ = > \frac{ \sqrt{2} ( \frac{3 - 10}{3} )}{ \sqrt{2}( \frac{4 + 24}{3} )} \\ \\ = > \frac{ - 7}{28} = \frac{ - 1}{4}$

I HOPE ITS HELP YOU DEAR,
THANKS