WHAT ARE OBLIQUE TRIANGLES? (WITH EXERCISES) - MATHS - 2024

The Oblique triangles are those triangles that are not rectangles. In other words, the triangles such that none of their angles is a right angle (their measure is 90º).

Since they have no right angles, then the Pythagorean Theorem cannot be applied to these triangles.

Therefore, to know the data in an oblique triangle it is necessary to use other formulas.

The formulas necessary to solve an oblique triangle are the so-called laws of sines and cosines, which will be described later.

In addition to these laws, the fact that the sum of the interior angles of a triangle is equal to 180º can always be used.

Oblique Triangles

As stated at the beginning, an oblique triangle is a triangle such that none of its angles measure 90º.

The problem of finding the lengths of the sides of an oblique triangle, as well as finding the measures of its angles, is called "solving oblique triangles."

An important fact when working with triangles is that the sum of the three internal angles of a triangle is equal to 180º. This is a general result, therefore for oblique triangles it can also be applied.

Laws of sines and cosines

Given a triangle ABC with sides of length "a", "b" and "c":

- The law of sines states that a / sin (A) = b / sin (B) = c / sin (C), where A, B and C are the opposite angles to «a», «b» and «c »Respectively.

- The law of cosines states that: c² = a² + b² - 2ab * cos (C). Equivalently, the following formulas can be used:

b² = a² + c² - 2ac * cos (B) or a² = b² + c² - 2bc * cos (A).

Using these formulas, the data for an oblique triangle can be calculated.

Exercises

Below are some exercises where the missing data of the given triangles must be found, based on certain data supplied.

First exercise

Given a triangle ABC such that A = 45º, B = 60º and a = 12cm, 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 that

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.

We can also solve for «c» and obtain that

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

Second Exercise

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, B = 180º-60º-75º = 45º. Furthermore, 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.

Third Exercise

Given 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 in the previous two 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 is then that

c² = 10² + 15² - 2 (10) (15) cos (80º) = 325 - 300 * 0.173 ≈ 272.905 cm, so that 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, we can obtain that A = 180º - 80º - 63.38º ≈ 36.62º.

Fourth Exercise

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 if we can apply the law of sines and thus obtain 5 / sin (A) = 3 / sin (B) = 7 / sin (120º), from where 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. Landaverde, F. d. (1997). Geometry (Reprint ed.). Progress.
2. Leake, D. (2006). Triangles (illustrated ed.). Heinemann-Raintree.
3. Pérez, CD (2006). Precalculation. Pearson Education.
4. Ruiz, Á., & Barrantes, H. (2006). Geometries. CR technology.
5. Sullivan, M. (1997). Precalculation. Pearson Education.
6. Sullivan, M. (1997). Trigonometry and Analytical Geometry. Pearson Education.