If cos 𝑥 equals one-half, find the value of 𝑥, where 𝑥 is greater than zero and less than 90 degrees.
We’ve been given information about the trigonometric ratio for cos of 𝑥. And we might recall that the cosine ratio tells us that, for a right triangle with angle 𝜃, the cos of 𝜃 is equal to the adjacent divided by the length of the hypotenuse. So, we need to find a right triangle such that the adjacent divided by the hypotenuse is equal to one-half. Now, in fact, the triangle that we used to generate this is an equilateral triangle of side length two units. The perpendicular bisector of one of the sides of this triangle will pass through the opposite vertex. So, we add the perpendicular bisector, creating a pair of congruent triangles of base length one unit.
So, what else do we know about this triangle? Well, we know the interior angles in an equilateral triangle are each 60 degrees. So, the angle at the vertex between the one-unit side and the two-unit side is 60 degrees. Then, since we bisected the angle at the other vertex, we have a 30-degree angle here. Since one-half is one divided by two, we need to find the angle 𝑥 such that the adjacent side to this angle is the one-unit side and the hypotenuse is the two-unit side.
We might notice that the one-unit side is in fact adjacent to the angle of 60 degrees, whilst the hypotenuse lies directly opposite the right angle, so it’s the two-unit length. We can therefore say that cos of 60 must be equal to one-half. Therefore, 𝑥 must be equal to 60 degrees. Now, whilst we derive this using the equilateral triangle of side length two units, this isn’t entirely necessary. In fact, we should learn that cos of 60 is equal to one-half, alongside the exact values for cos of zero, cos of 30, cos of 45, and cos of 90 degrees.
Using either method, we find that the value of 𝑥 such that cos of 𝑥 is equal to one-half and 𝑥 is in the open interval from zero to 90 degrees is 60.