Trigonometry

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THE 30°-60°-90° TRIANGLE

THERE 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.

Theorem.  In a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : square root of 3.

30-60-90 triangle

We will prove that below.

(For the definition of measuring angles by "degrees," see Topic 12.)

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 square root of 3.  Also, while 1 : square root of 3: 2 correctly corresponds to the sides opposite 30°-60°-90°, many find the sequence 1 : 2 : square root of 3 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°.

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.

30-60-90 triangle

Since the cosine is the ratio of the adjacent side to the hypotenuse, you can see that cos 60° = ½.

Example 2.   Evaluate sin 30°.

Answer.   According to the property of cofunctions (Topic 3), sin 30° is equal to cos 60°.   sin 30° = ½.

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.
To cover the answer again, click "Refresh" ("Reload").

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

    sin 60° =  square root of 3
 2
 = ½square root of 3.

Lesson 5 of Algebra.

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

    tan 60° =  square root of 3
 1
= square root of 3.

Problem 2.   Evaluate cot 30° and cos 30°.

1 : 2: SqRt 3

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

   Therefore, on inspecting the figure above, cot 30° =  square root of 3
 1
 = square root of 3.

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° =  square root of 3
 2
 = ½square root of 3.

Before we come to the next Example, here is how we relate the sides and angles of a triangle:

Triangle ABC

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.

1:2:SqRt3

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 : square root of 3, as shown on the right. Whenever we know the ratios of the sides, we can solve the triangle by the method of similar figures.

And so 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 5square root of 3cm.

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 5square root of 3cm.

Whenever we know the ratio numbers, the student should use this method of similar figures to solve the triangle, and not the trigonometric Table.

(In Topic 6, we will solve right triangles the ratios of whose 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 d and f ?

1 : 2: Sq rt 3

1 : 2: SqRt 3

   The student should draw a similar triangle in the same orientation. Then see that the side corresponding to square root of 3 was multiplied by square root of 3. Therefore, each side will be multiplied by square root of 3. Side d will be 1timessquare root of 3 = square root of 3. Side f will be 2square root of 3.

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

1 : 2: SqRt 3

1 : 2: SqRt 3

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 ½square root of 3.

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

30-60-90 triangle

30-60-90 triangle

The side adjacent to 60° is always half of the hypotenuse -- therefore, side b is 9.3 cm.
But this is the side that corresponds to 1. And it has been multiplied by 9.3.  Therefore, side a will be multiplied by 9.3.
It will be 9.3square root of 3 cm.

Example 4.   ABC is an equilateral triangle whose height AD is 4 cm.

30-60-90 triangle

Find the length of the side x.

Solution 1.   Since the triangle is equilateral, it is also equiangular, and therefore the the angle at B is 60°

The height of a triangle is the straight line drawn from the vertex at right angles to the base.  Therefore, triangle ADB is a 30-60-90 triangle.

30-60-90 triangle

For this problem, it will be convenient to form the proportion with fractional symbols:

x
4
 =   2 
square root of 3

That implies

x  =  4· 2
square root of 3
 =   8 
square root of 3
 .

On taking square root of 3to be approximately 1.732,

 _8_ 
1.732
approximately 4.619 cm.
  Solution 2. 30-60-90 triangle

The side corresponding to square root of 3 was multiplied to become 4.  How was it multiplied?

By    4 
square root of 3
:
 4 
square root of 3
·  square root of 3   =  4.

Compare the previous topic.

Therefore the hypotenuse 2 will also be multiplied by   4 
square root of 3
:
 4 
square root of 3
·  2   =    8 
square root of 3
,

which, again, is approximately 4.619 cm.

Problem 6.   Inspect the values of 30°, 60°, and 45° -- that is, look at the two triangles --

30-60-90 triangle

-- and in each equation, decide which of those angles is the value of x.

   a)   sin x = cos xx = 45°.   b)   tan x = 1.  x = 45°.  
 
   c)   sin x = ½.   x = 30°   d)   cos x = ½.  x = 60°.
 
   e)   sin x = ½square root of 3.   x = 60°   f)   cos x = ½square root of 2x = 45°.
  30-60-90 triangle   x = 30°.   h)   csc x = 2.  x = 30°.

Example 5.   Solve this equation for angle x:

  2 sin (x − 10°) − square root of 3 = 0.
 
 Solution. 2 sin (x − 10°) − square root of 3 = 0
 
implies  
  sin (x − 10°) = ½square root of 3.

Now, the sine of what angle is ½square root of 3?

  sin 60° = ½square root of 3.
 
Therefore,  
  x − 10° = 60°.
 
  x = 70°.

Problem 7.   Solve this equation for angle x:

  3 csc (2x + 6°) − 6 = 0.
 
  3 csc (2x + 6°) − 6 = 0
 
implies  
  csc (2x + 6°) = 2.
 
  2x + 6° = 30°
 
  x + 3° = 15°
 
  x = 12°.

Problem 8.    Prove:  The area A of an equilateral triangle whose side is s, is

A = ¼square root of 3s2.

30-60-90 triangle

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. (Topic 2, Problem 6.)

In an equilateral triangle each side is s , and each angle is 60°.  Therefore,

A = ½ sin 60°s2.

Since sin 60° = ½square root of 3 (Problem 1),

A = ½· ½square root of 3 s2 = ¼square root of 3s2.

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

A 3
4
square root of 3 r2.

30-60-90 triangle

30-60-90 triangle

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° = ½square root of 3.

Therefore,

s = square root of 3r

so that

s2 = 3r2.

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

A = ¼square root of 3s2.

Therefore,

A = ¼square root of 3s2 = ¼square root of 3· 3r2  =  3
4
square root of 3r2.

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.

30-60-90 triangle

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.

30-60-90 triangle

  Now,  BP
PD
= csc 30° =  2.

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.

The proof

Here is the proof that in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : square root of 3.  It is based on the fact that a 30°-60°-90° triangle is half of an equilateral triangle.

Draw the equilateral triangle ABC.  Then each of its equal angles is 60°.  (Theorems 3 and 9)

A 30-60-90 triangle

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 = square root of 3.

Therefore in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : square root of 3; which is what we set out to prove.


Next Topic:  The Isosceles Right Triangle


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