The hypotenuse of,c of right triangle ABC is 7.0 m long. A trigonometric ratio for angle A is given for four different triangles. Which one of these triangles has the greatest area? Justify your decision. a)sec A=1.7105 b)cos A=0.7512 c) csc A=2.2703 d)sin A= 0.1515

Answer 1

To visualize the problem:

We are given the hypotenuse of a right triangle, i.e. the angle opposite the 7m side is 90°.

Recall from elementary geometry that a triangle with the diameter as the base, and the third point anywhere on the circle is a right triangle.

So the given base of 7m in length can be considered as the diameter of a circle. Try to place a point on the circle to make the height a maximum.

The point is evidently when angles A and B both equal 45°.

In fact, the height, h, is given by:
h=c cos(A) sin(A)
=(c/2)sin(2A) [since sin(2A)=2sin(A)cos(A)]
Therefore the maximum area happens when sin(2A) has the greatest value.
a) secA=1.7105, A=54.22°, sin(2A)=0.8898
b) cosA=0.7512, A=41.3°, sin(2A)=0.9917
c) cscA=2.2703, A=26.13° sin(2A)=0.7908
d) sinA=0.1515, A=8.714°, sin(2A)=0.2995

I will leave it to you to choose the largest value of sin(2A), which results in the highest value of area, where
A=(1/2)c*(c/2)sin(2A)
=(c/2)² sin(2A)

Answer 2

Thanks!