# Oblique Triangles: Characteristics, Examples, Exercises

The Oblique triangles  are those who have no right angles, so I any of its internal angles is equal to 90 °. So an oblique triangle can be either acute or obtuse .

In the first case, the internal angles of the triangle are acute or what is the same: less than 90º, while in the second, there is always an angle greater than 90º, that is, an obtuse angle. Let’s see an example of each in the following figure: To find the lengths of the sides and the measures of the interior angles of this class of triangles, in the absence of right angles it is not possible to apply the Pythagorean theorem.

However, there are alternatives to solve the triangle: the cosine and sine theorems and the fact that the sum of the interior angles is equal to 180º.

Article index

• one

Examples of oblique triangles

• 1.1

Acute triangle

• 1.2

Obtuse triangle

• two

Theorems of the sinus and cosine

• 2.1

Sine theorem

• 2.2

Cosine theorem

• 3

Solved exercises

• 3.1

– Exercise 1

• 3.2

– Exercise 2

• 3.3

– Exercise 3

• 3.4

– Exercise 4

• 3.5

– Exercise 5

• 3.6

– Exercise 6

• 4

References

## Examples of oblique triangles

Guiding ourselves by figure 1, we can easily recognize oblique triangles by two criteria that we will give below.

### Acute triangle

Let the triangle with sides a, b and c, with α be the angle opposite side a.

If the square on side a opposite the acute angle α is less than the sum of the squares on the remaining sides, the triangle is acute . Algebraically:

a 2 <b 2 + c 2 ; α <90º

The familiar equilateral triangle, one that has its three sides of the same measure, is acute and therefore oblique, since its internal angles are equal and measure 60º.

### Obtuse triangle

On the other hand, if the square of the side opposite to the obtuse angle α is greater than the sum of the squares of the other two, we are in the presence of an obtuse triangle . Therefore:

a 2 > b 2 + c 2 ; α> 90º

For example, a triangle whose interior angles are 105º, 60º, and 15º is an obtuse oblique triangle. Note that 105º + 60º + 15º = 180º.

## Theorems of the sinus and cosine

To solve oblique triangles, that is, find the measures of all their sides and all their angles, the theorems of sine and cosine are required.

Let a, b and c be the sides of a triangle, and α, β and γ its internal angles. Then:

### Sine theorem

The sine theorem states the following:

Where α is the angle opposite side a, β is the angle opposite side b and γ is the angle opposite side c.

Equivalently:

We choose to apply the sine theorem when we are going to solve a triangle for which we know more angles than sides.

### Cosine theorem

According to the cosine theorem:

c 2 = a 2 + b 2 – 2⋅a⋅b⋅cos γ

Again the angle γ is in front of the side c. Likewise, we can write equivalent expressions for sides a and b, as follows:

a 2 = b 2 + c 2 – 2⋅b⋅c⋅cos α

Y

b 2 = a 2 + c 2 – 2⋅a⋅c⋅cos β

The cosine theorem is preferably applied when the value of two sides and the angle between them are known. Likewise, once the three sides of a triangle are known, the theorem allows us to calculate the cosine of the angle between two of them.

## Solved exercises ### – Exercise 1

Check that the triangle whose sides measure 20, 10 and 12 arbitrary units is obtuse.

#### Solution

We do not know any of the internal angles, but taking into account the criterion used to recognize obtuse triangles, we can pose the inequalities with the squares of the sides to see if it is true.

First we find the squares on each side:

20 2 = 400

10 2 = 100

12 2 = 144

And we see that in effect: 400> 100 + 144, since 400> 244. Therefore, the triangle contains an angle greater than 90º, located opposite the side that measures 20. Consequently, this triangle, in addition to being oblique, it is also obtuse.

### – Exercise 2

Given the oblique triangle shown in Figure 2, whose measurements are given in arbitrary units, determine:

a) The value of x. Is it an acute or an obtuse triangle?

b) The remaining interior angles of the triangle

c) Perimeter

d) Area. #### Solution to

Two adjacent sides of the triangle are known, whose measures are 38.0 and 45.8 and the angle between them, which is 30º, therefore the cosine theorem is of immediate application:

x 2 = 38.0 2 + 45.8 2 – 2 x 38.0 x 45.8 x cos 30º = 527.18

Therefore:

x = (527.18) 1/2 = 22.96

The drawing suggests that α> 90º and the triangle is obtuse as well as oblique. To check this, we find the squares of the sides, as was done in the previous exercise:

22.96 2 = 527.18

38.0 2 = 1444.00

45.8 2 = 2097.64

The angle α is greater than 90º if the square of the opposite side is satisfied: 45.8 2   is greater than the sum of the squares of the other sides, which is 22.96 2 + 38.0 2 .

Let’s see if it happens like this:

527.18 + 1444.00 = 1971.2

Indeed:

2097.64> 1971.2

Therefore the angle α is greater than 90º.

#### Solution b

Now we can apply the sine theorem to find one of the missing angles. Let’s propose it for the angle β:

sin 30º / 22.96 = sin β / 38

sin β = 38 x (sin 30º / 22.96) = 0.8275

β = arcsen (0.8275) = 55.84º

We can find the missing angle knowing that the sum of the internal angles of any triangle is 180º. Therefore:

55.84º + 30º + α = 180º

α = 94.16º

If you prefer, you can also use the cosine theorem to find the cosine of the angle between any two adjacent sides. Once obtained, the arc cosine function is used to determine the angle.

The results may differ slightly to the decimal places, depending on the rounding carried out.

#### Solution c

The perimeter P is the contour of the figure, equivalent to the sum of the measurements of the three sides:

P = 22.96 + 38.00 + 45.80 = 106.76 arbitrary units.

#### Solution d

The formula to calculate the area of ​​any triangle is:

A = (1/2) x base x height

We need to choose one of the sides as the base and determine the height. For example, choosing the side that measures 45.8, we draw the height h to vertex A, which is the red line in Figure 2b.

By doing this we divide the original triangle into two right triangles, both with h as a common leg. Either one works, since we know a side and an acute angle.

We are going to take the one with a hypotenuse equal to 38, a leg that measures h, which is the height sought and the acute angle equal to 30º.

With the help of the trigonometric ratios of the acute angle 30º we determine the value of h :

sin 30º = leg opposite 30º / hypotenuse = h / 38

h = 38 x sin 30º = 19

Therefore:

A = (1/2) x 45.8 x 19 = 435.1 arbitrary units of area.

We could have chosen another of the sides as the base, for example side 38, in this case, the height h is different, since another right triangle is formed, but the result of the area is the same. It remains as an exercise for the reader to verify it.

### – Exercise 3

Given a triangle ABC such that A = 45º, B = 60º and a = 12 cm, calculate the other data of the triangle.

#### Solution

Using that the sum of the internal angles of a triangle is equal to 180º we have:

C = 180º-45º-60º = 75º.

The three angles are already known. The law of sines is then used to calculate the two missing sides.

The equations that arise are 12 / sin (45º) = b / sin (60º) = c / sin (75º).

From the first equality we can solve for “b” and obtain that:

b = 12 * sin (60º) / sin (45º) = 6√6 ≈ 14.696cm.

You can also solve for “c” and obtain that:

c = 12 * sin (75º) / sin (45º) = 6 (1 + √3) ≈ 16.392cm. ### – Exercise 4

Given triangle ABC such that A = 60º, C = 75º and b = 10cm, calculate the other data of the triangle.

#### Solution

As in the previous exercise, we have that B = 180º-60º-75º = 45º. In addition, using the law of sines, we have that a / sin (60º) = 10 / sin (45º) = c / sin (75º), from which it is obtained that a = 10 * sin (60º) / sin (45º) = 5√6 ≈ 12.247 cm and c = 10 * sin (75º) / sin (45º) = 5 (1 + √3) ≈ 13.660 cm. ### – Exercise 5

Given the triangle ABC such that a = 10cm, b = 15cm and C = 80º, calculate the other data of the triangle.

#### Solution

In this exercise, only one angle is known, therefore it cannot be started as it was done in the two previous exercises. Also, the law of sines cannot be applied because no equation could be solved.

Therefore, we proceed to apply the law of cosines. It then follows that:

c² = 10² + 15² – 2 (10) (15) cos (80º) = 325 – 300 * 0.173 ≈ 272.905 cm,

So c ≈ 16.51 cm. Now, knowing the 3 sides, the law of sines is used and it is obtained that:

10 / sin (A) = 15 / sin (B) = 16.51cm / sin (80º).

Hence, solving for B results in sin (B) = 15 * sin (80º) / 16.51 ≈ 0.894, which implies that B ≈ 63.38º.

Now, it can be obtained that A = 180º – 80º – 63.38º ≈ 36.62º. ### – Exercise 6

The sides of an oblique triangle are a = 5cm, b = 3cm, and c = 7cm. Find the angles of the triangle.

#### Solution

Again, the law of sines cannot be applied directly, since no equation would serve to obtain the value of the angles.

Using the cosine law we have that c² = a² + b² – 2ab cos (C), from which when solving we have that cos (C) = (a² + b² – c²) / 2ab = (5² + 3²-7²) / 2 * 5 * 3 = -15/30 = -1/2 and therefore C = 120º.

Now we can apply the law of sines and thus obtain 5 / sin (A) = 3 / sin (B) = 7 / sin (120º), from which we can solve for B and obtain that sin (B) = 3 * sin (120º) / 7 = 0.371, so that B = 21.79º.

Finally, the last angle is calculated using that A = 180º-120º-21.79º = 38.21º. ## References

1. Clemens, S. Geometry with Applications. Addison Wesley.
2. Ibáñez, P. 2010. Mathematics III. Cengage Learning.
3. Jiménez, R. Mathematics II: Geometry and Trigonometry. 2nd. Edition. Pearson.
4. Math For You. Obtuse triangle. Recovered from: matematicasparaticharito.wordpress.com.
5. Stewart, J. 2007. Precalculus. 5th. Edition. Cengage Learning.