Question

guys please help me out I'm having a headache with this​

Answer 1

Answer:

17/19

Step-by-step explanation:

Let y=(sinA + cosA)/(2cosA-sinA)

Taking CosA as common from both numerator and denominator

y=[cosA((sinA/cosA)+(cosA/cosA))] /[cosA(2(cosA/cosA)-(sinA/cosA))]

Cancelling cosA and as we know sinA/cosA=tanA

So,

y=(tanA+1)/(2-tanA)

And we know

5cosA-12sinA=0

5cosA =12sinA

5=12[(sinA/cosA)]

TanA=5/12

Substitute the value of tanA in y

we get

y=(1+(5/12))/(2-(5/12))

y=[(12+5)/12]/[(12*2–5)/12]

Cancelling 12

y=17/(24–5)

y=17/19

Answer 2

Step-by-step explanation:

5cos-12sin=0

5cos=12sin

cos/sin= 12/5

cot= 12/5

b/p=12/5

hence b= 12 and p=5

h²=12²+5²

h²= 144+25

h²= 169

h= 13

(sin+cos)/2cos-sin

(p/h+b/h)/(2b/h -p/h)

{(p+b)/h}/{(2b-p)/h)

(p+b)/(2b-p)

(12+5)/(2•12-5)

17/(24-5)

17/19

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