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