If θ lies in the first quadrant and \(\cot \theta = \frac{{63}}{{16}}\) , then what is the value of (sin θ + cos θ)?
Given:
\(\cot θ = \frac{{63}}{{16}}\)
Formula Used:
sin θ = Perpendicular/Hypotenus
cos θ = Base/ Hypotenuse
cot θ = Perpenicular/Base
Pythagorus theorem
(Hypotenus) 2= (Perpendicular) 2+ (Base) 2
Calculation:
\(\cot θ = \frac{{63}}{{16}}\) = B/P
Here, hypotenuse = \(\sqrt {{{63}^2} + {{16}^2}} = \sqrt {4225} \) = 65
So,
sinθ = P/H = 16/65
cosθ = B/H = 63/65
∴ (sin θ + cos θ) = (16/65) + (63/65) = 79/65Similar Questions
If sinθ = (m 2– n 2)/(m 2+ n 2) and 0 < θ < π/2, then what is the value of cosθ?
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