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7

THE 30°-60°-90° TRIANGLE

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 .  And while 1::2 more correctly corresponds to 30°-60°-90°, many find the sequence 1:2: easier to remember.)

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°.

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°.

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The sine is the ratio of the opposite side to the hypotenuse.

sin 60° =

2

= ½.

The tangent is ratio of the opposite side to the adjacent.

tan 60° =

1

= .

The cotangent is the ratio of the adjacent side to the opposite.

Therefore, on inspecting the figure above, cot 30° =

1

= .

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,

cos 30° =

2

= ½.

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 : , as shown on the right. Whenever we know the ratios of the sides, we can solve the triangle by the method of similar figures.

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 cm.

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 cm.

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 ?

Problem 4.   In the right triangle PQR, angle P is 30°, and side r is 1 cm.  How long are sides p and q ?

Problem 5.   Solve the right triangle ABC if angle A is 60°, and the hypotenuse is 18.6 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.

a)	sin x = cos x.  x = 45°.		b)	tan x = 1.  x = 45°.
c)	sin x = ½.   x = 30°		d)	cos x = ½.  x = 60°.
e)	sin x = ½.   x = 60°		f)	cos x = ½.  x = 45°.

x = 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.
And the center of a circle lies on the perpendicular bisector of any chord (Theorem 11);
therefore AD passes through the center O.
Hence AO, OB are radii, and triangle AOB is isosceles.
Therefore angles OAB, ABO are equal (Theorem 3):
angle ABO is also 30°.
Therefore, since angle ABD is 60°, angle OBD is 30°;
which is what we wanted to prove.

b)  Prove:  The area A of an equilateral triangle inscribed in a circle of radius r, is

A =

3 4

r².

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.
Therefore,

½s  r

=

cos 30° = ½.

Hence,

s = r

so that

s² = 3r².

Now, according to the previous problem, the area A of an equilateral triangle is

A = ¼s².

Therefore,

A = ¼s² = ¼· 3r²

=

3 4

r².

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.
Then AD is the perpendicular bisector of BC  (Theorem 2).  Triangle ABD therefore is a 30°-60°-90° triangle.

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:

AD² + 1²	=	2²
AD²	=	4 − 1 = 3
AD	=	.

Therefore in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : ; which is what we set out to prove.

Next Topic:  Solve Right Triangles

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