Math, asked by Denniskhiangte7317, 10 months ago

In right Δ ,∟=90°, BC= 5cm and AC= 13cm,then evaluate : 1+sinC/1+cosC

Answers

Answered by VivekTuli
0

Answer:

25/18

if AB =P =√13^2-5^2 = 12 cm and BC =B=5 cm and AC = H= 13cm

Step-by-step explanation:

then sinC = P/H=AB/AC=12/13

cosC = B/H=BC/AC =5/13

1+sinC/1+cosC = 1+(12/13)/1+(5/13)=13+12/13+5 =25/18

Answered by ITzBrainlyGuy
2

CORRECT QUESTION:

In a right angled triangle ABC right angle at B , BC = 5cm & AC = 13cm

Find 1 + sinC/1 + cosC

ANSWER:

Given

  • ∆ABC is a right angled triangle
  • Right angle at B
  • BC = 5cm , AC = 13cm , AB = ?

Here AC is hypotenuse , BC is base & AB is height

\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){30}}\put(30,0){\line(0,1){40}}\put(0,0){\line(3,4){30}}\put(13,-4){5\ cm}\put(32,17){12\ cm}\put(8,23){13 cm}\put(3,1){$\theta$}\end{picture}

Using Pythagoras theorem

AC² = AB² + BC²

13² = AB² + 5²

169 - 25 = AB²

AB = √144

AB = 12cm

Let us consider angle C as ‘θ’

sinθ = opposite/hypotenuse

sinθ = 12/13

cosθ = adjacent/hypotenuse

cosθ = 5/13

Substitute we get

♦ 1 + 12/13/1 + 5/13

♦ 25/13/18/13

♦ 25/18

Hence, 1 + sinC/1 + cosC = 25/18

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