If #sinalpha=4/5# and #alpha# is obtuse angle, what is #cosalpha#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Mar 2, 2017 #cosalpha=-3/5# Explanation: As #alpha# is obtuse, it lies in second quadrant and #cosalpha# will be negative. Now #sinalpha=4/5# and therefore #cosalpha=-sqrt(1-(4/5)^2)=-sqrt(1-16/25)=-sqrt(9/25)=-3/5# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 2518 views around the world You can reuse this answer Creative Commons License