If sinA=3/5,then find cosC+sinC. Please give answer
Answers
Answered by
1
Here given sin A = 3/5
We know Sin = p/h ( Where p = perpendicular and h = hypotenuse).
So, For angle A,p = 3k and h = 5k.
Let the base be b.
So, By applying Pythagoras Theorem, we get
p^2 + b^2 = h^2
9k^2 + b^2 = 25k^2
b^2 = 16k^2
b = 4k.
Now, For angle C , b = 3k , p = 4k and h = 5k.
cosC + sinC
= 3k/5k + 4k/5k
= 3/5 + 4/5
= (3+4)/5
= 7/5
We know Sin = p/h ( Where p = perpendicular and h = hypotenuse).
So, For angle A,p = 3k and h = 5k.
Let the base be b.
So, By applying Pythagoras Theorem, we get
p^2 + b^2 = h^2
9k^2 + b^2 = 25k^2
b^2 = 16k^2
b = 4k.
Now, For angle C , b = 3k , p = 4k and h = 5k.
cosC + sinC
= 3k/5k + 4k/5k
= 3/5 + 4/5
= (3+4)/5
= 7/5
yashagarwalla001:
was it helpful
wc
yes it helpful
wc again
thanks a lot!!☺️
yo
Answered by
0
Answer:
Step-by-step explanation:
Similar questions