![]() 7 A 30°-60°-90° TRIANGLE is another standard mathematical object. The student should know the ratios of the sides. Theorem. In a 30°-60°-90° triangle the sides are in the ratio
1:2: 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 The cited theorems are from the Appendix, Some theorems of plane geometry. Before giving the proof for those ratios, here are some examples of how we take advantage of knowing them. First, we can evaluate the functions of 60° and 30°. Example 1. Evaluate cos 60°. Answer. For any problem involving a 30°-60°-90° triangle, the student should not use a table. The student should sketch the triangle and place the ratio numbers. Since the cosine is the ratio of the adjacent side to the hypotenuse, we see that cos 60° = ½. Example 2. Evaluate sin 30°. Answer. According to the property of cofunctions (Topic 5), sin 30° is equal to cos 60°. sin 30° = ½. The student can also 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°. Problem 1. 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 a. Similarly for angle B and side b, angle C and side c. Example 3. Solve the right triangle ABC if angle A is 60°, and side c is 10 cm. ![]() Solution. To solve a triangle means to know all three sides and all three angles. Since this is a right triangle and angle A is 60°, then the remaining angle B is its complement, 30°. Now in every 30°-60°-90° triangle, the sides are in the ratio 1 : 2 : Thus, in triangle ABC, the side corresponding to 2 has been multiplied by 5. Therefore every side will be multiplied by 5. Side b will be 5 × 1, or simply 5 cm, and side a will be 5 Alternatively, we could say that the side adjacent to 60° is always half of the hypotenuse. Therefore, side b will be 5 cm. Now, side b is the side that corresponds to 1. And it has been multiplied by 5. Therefore, side a must also be multiplied by 5. It will be 5 Whenever we know the ratio numbers, we use this method of similar figures to solve the triangle, and not the trigonometric Table. (In Topic 8, we will solve right triangles the ratios of whose sides we do not know.) Problem 3. In the right triangle DEF, angle D is 30°, and side DF is 3 inches. How long are sides d and f ? ![]()
The student should draw a similar triangle in the same orientation. We then see that the side corresponding to Problem 4. In the right triangle PQR, angle P is 30°, and side r is 1 cm. How long are sides p and q ? ![]()
The side corresponding to 2 has been divided by 2. Therefore, each side must be divided by 2. Side p will be ½, and side q will be ½ Problem 5. Solve the right triangle ABC if angle A is 60°, and the hypotenuse is 18.6 in.
The side adjacent to 60° is always half of the hypotenuse -- therefore, side b is 9.3 in. Problem 6. Inspect the values of 30°, 60°, and 45° -- that is, look at the two triangles -- and decide which of those angles satisfies each equation.
Problem 7. Prove: The area A of an equilateral triangle whose side is s, is A = ¼ The area A of any triangle is equal to one-half the sine of any angle times the product of the two sides that make the angle. (Lesson 4, Problem 5.) In an equilateral triangle each side is s , and each angle is 60°. Therefore, A = ½ sin 60°s². Since sin 60° = ½ A = ½· ½ a) ABC is an equilateral triangle inscribed in a circle with center O. a) AD bisects the angle at A. Prove that angle OBD is 30°.
The straight line that bisects the vertex angle of an isosceles triangle is the perpendicular bisector of the base (Theorem 2); therefore AD is the perpendicular bisector of the base CB. b) Prove: The area A of an equilateral triangle inscribed in a circle of radius r, is
Let AD be the perpendicular bisector of side CB. Let O be the center of the circle with radius OB, and call the side of the inscribed equilateral triangle s. Then according to part a), triangle BOD is a 30-60-90 triangle.
Hence, s = so that s² = 3r². Now, according to the previous problem, the area A of an equilateral triangle is A = ¼ Therefore,
The proof Here is the proof that in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : 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 : Next Topic: Solve Right Triangles Please make a donation to keep TheMathPage online. Copyright © 2001-2007 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |