Identify the triangle that contains an acute angle for which the sine and cosine ratios are equal.

Solution:

A right-angled triangle is one in which one angle is 90°. One such triangle is shown below:

The above right-angled triangle is an isosceles triangle with two sides AB and BC being equal. If AB = x then BC is also = x.

Then AC is calculated using the Pythagorean triplet theorem which states;

AC² = AB² + BC² (from Pythagoras Theorem)

AC² = x² + x²

AC² = 2x²

AC = x√2

The trigonometric ratios for the above right angled triangle will be as follows:

sin 45° = Perpendicular/Hypotenuse = AB/AC = x/x√2 = 1/√2

cos 45° = Base/Hypotenuse = x/x√2 = 1/√2

tan 45° = Perpendicular/Base = Sin 45°/Cos 45° = 1/√2/1/√2 = 1

Hence the second triangle in the statement problem is the correct triangle in which the angles are 90°, 45°, and 45°.

Identify the triangle that contains an acute angle for which the sine and cosine ratios are equal.

Summary:

The triangle that contains an acute angle for which the sine and cosine ratios are equal is the triangle with angles 90°, 45°, and 45°.