SOLUTION: The hypotenuse, c, of right angle ABC is 10.2 cm long. Given the trigonometric ratio sec(A) = 15/3 for angle A, what is the area of the triangle to the nearest tenth of a cm^2?

Algebra ->  Trigonometry-basics -> SOLUTION: The hypotenuse, c, of right angle ABC is 10.2 cm long. Given the trigonometric ratio sec(A) = 15/3 for angle A, what is the area of the triangle to the nearest tenth of a cm^2?      Log On


   

Question 1201967: The hypotenuse, c, of right angle ABC is 10.2 cm long. Given the trigonometric ratio sec(A) = 15/3 for angle A, what is the area of the triangle to the nearest tenth of a cm^2?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

The hypotenuse, c, of right angle ABC is 10.2 cm long. Given the trigonometric ratio sec(A) = 15/3 for angle A, what is the area of the triangle to the nearest tenth of a cm^2?
Sec A =15/3 =5:1
cos A = AB/AC =1/5

1/5 = AB/10.2
AB= 2.04 cm
BC calculate by Pythagoras Theorem
sqrt(10.2^2-2.04^2)= 9.999
Area of triangle = 1/2 *B * h
1/2 * 9.999 *2.04 = 10.198 cm^2 ( you round off)
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Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
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Why in your post do you represent the number of 5 (the value of secant) as 15/3 ?

Is your goal to scare people around ?