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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°.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

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 cm.

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°.

Example 4.   Solve this equation for angle x:

	2 sin (x − 10°) −	=	0.
Solution.	2 sin (x − 10°) −	=	0
implies
	sin (x − 10°)	=	½.

Now, the sine of what angle is ½?

	sin 60°	=	½.
Therefore,
	x − 10°	=	60°.
	x	=	70°.

Problem 7.   Solve this equation for angle x:

Problem 9.   Prove:  The area A of an equilateral triangle inscribed in a circle of radius r, is

A =

3 4

r².

The three radii divide the triangle into three congruent triangles (Side-side-side); hence each radius bisects each vertex into two 30° angles.
If we extend the radius AO, then AD is the perpendicular bisector of the side CB. (Theorem 2.) Triangle OBD is therefore a 30-60-90 triangle.
If we call each side of the equilateral triangle s, then in the right triangle OBD,

½s  r

=

cos 30° = ½.

Therefore,

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.

That is what we wanted to prove.

Problem 10.   Prove:  The angle bisectors of an equilateral triangle meet at a point that is two thirds of the distance from the vertex of the triangle to the base.

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°;
and the side BD is equal to the side AE, because in an equilateral triangle the angle bisector is the perpendicular bisector of the base (Theorem 2);
angles PDB, AEP then are right angles and equal.
Therefore triangles BPD, APE are congruent: Angle-side-angle.

Therefore, BP = 2PD.

But AP = BP, because triangles APE, BPD are conguent, and those are the sides opposite the equal angles.
Therefore, AP = 2PD.
     Therefore AP is two thirds of the whole AD.
Which is what we wanted to prove.

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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