8 ## THE 30°-60°-90° TRIANGLETHERE ARE TWO special triangles in trigonometry. One is the 30°-60°-90° triangle. The other is the isosceles right triangle. They are special because with simple geometry we can know the ratios of their sides, and therefore solve any such triangle. Theorem. We will prove that below. Note that the smallest side, 1, is opposite the smallest angle, 30°; while the largest side, 2, is opposite the largest angle, 90°. (Theorem 6). (For, 2 is larger than . Also, while 1 : : 2 correctly corresponds to the sides opposite 30°-60°-90°, many find the sequence 1 : 2 : easier to remember.) The cited theorems are from the Appendix, Some theorems of plane geometry. Here are examples of how we take advantage of knowing those ratios. First, we can evaluate the functions of 60° and 30°. Example 1. Evaluate cos 60°.
Since the cosine is the ratio of the adjacent side to the hypotenuse, we can see that cos 60° = ½. Example 2. Evaluate sin 30°.
On the other hand, you can see that directly in the figure above. Problem 1. Evaluate sin 60° and tan 60°. To see the answer, pass your mouse over the colored area. The sine is the ratio of the opposite side to the hypotenuse.
The tangent is ratio of the opposite side to the adjacent.
Problem 2. Evaluate cot 30° and cos 30°. The cotangent is the ratio of the adjacent side to the opposite.
= . Or, more simply, cot 30° = tan 60°. As for the cosine, it is the ratio of the adjacent side to the hypotenuse. Therefore,
Before we come to the next Example, here is how we relate the sides and angles of a triangle: If an angle is labeled capital A, then the side opposite will be labeled small Example 3. Solve the right triangle ABC if angle A is 60°, and side AB is 10 cm.
Again, in every 30°-60°-90° triangle, the sides are in the ratio 1 : 2 : , as shown on the left. When we know the ratios of the sides, then to solve a triangle we do not require the trigonometric functions or the Pythagorean theorem. We can solve it by the method of similar figures. Now, the sides that make the equal angles are in the same ratio. Proportionally, 2 : 1 = 10 : AC. 2 is two times 1. Therefore 10 is two times AC. AC is 5 cm. The side adjacent to 60°, we see, is always As for BC—proportionally, 2 : = 10 : BC. To produce 10, 2 has been multiplied by 5. Therefore, will also be multiplied by 5. BC is 5 cm. In other words, since one side of the standard triangle has been multiplied by 5, then every side will be multiplied by 5. 1 : 2 : = 5 : 10 : 5. Compare Example 11 here. Again: When we know the ratio numbers, then to solve the triangle the student should use this method of similar figures, not the trigonometric functions. (In Topic 10, we will solve right triangles whose ratios of sides we do not know.)
Problem 3. In the right triangle DFE, angle D is 30° and side DF is 3 inches. How long are sides The student should draw a similar triangle in the same orientation. Then see that the side corresponding to was multiplied by .
Therefore, each side will be multiplied by . Side
Problem 4. In the right triangle PQR, angle P is 30°, and side
The side corresponding to 2 has been Problem 5. Solve the right triangle ABC if angle A is 60°, and the hypotenuse is 18.6 cm.
The side adjacent to 60° is always
Problem 6. Prove:
The area In an equilateral triangle each side is
Since sin 60° = ½,
Problem 7. Prove:
The three radii divide the triangle into three congruent triangles. Hence each radius bisects each vertex into two 30° angles. Triangle OBD is therefore a 30-60-90 triangle.
Therefore,
so that
Now, the area
Therefore,
That is what we wanted to prove.
Problem 8. Prove: Let ABC be an equilateral triangle, let AD, BF, CE be the angle bisectors of angles A, B, C respectively; then those angle bisectors meet at the point P such that AP is two thirds of AD. First, triangles BPD, APE are congruent.
For, since the triangle is equilateral and BF, AD are the angle bisectors, then angles PBD, PAE are equal and each
30°; Angles PDB, AEP then are right angles and equal. Therefore, triangles BPD, APE are congruent.
Therefore, BP = 2PD.
But AP = BP, because triangles APE, BPD are conguent, and those are the sides opposite the equal angles. The proof Here is the proof that in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : . It is based on the fact that a 30°-60°-90° triangle is Draw the equilateral triangle ABC. Then each of its equal angles is 60°. (Theorems 3 and 9) Draw the straight line AD bisecting the angle at A into two 30° angles. Now, since BD is equal to DC, then BD is half of BC. This implies that BD is also half of AB, because AB is equal to BC. That is, BD : AB = 1 : 2 From the Pythagorean theorem, we can find the third side AD:
Therefore in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : ; which is what we set out to prove. Corollary. The square drawn on the height of an equalateral triangle is three fourths of the square drawn on the side. Next Topic: The Isosceles Right Triangle Please make a donation to keep TheMathPage online. Copyright © 2022 Lawrence Spector Questions or comments? E-mail: [email protected] |