<PRE><B>Question 1201967</B>

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)

*[illustration RTriangle.png].


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