Precalculus

Trigonometry

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Solving Right Triangles

An arbitrary triangle is defined by three components, at least one of which must be a side. For example, a triangle can be defined by three sides \(\left({SSS}\right),\) or by a side and two adjacent angles \(\left({ASA}\right).\)

In a right triangle, one of the angles is equal to \(90^\circ.\) Therefore, to find all sides and angles of a right triangle, it is enough to know only two elements, at least one of which must be a side.

There are \(5\) basic combinations of sides and angles that uniquely define a right triangle:

Solving a triangle means finding all its sides and angles. Under finding an angle we mean finding a trigonometric function of the angle.

Depending on the given data, we use different relationships and identities. Let's consider these scenarios in more detail.

1. Solving a Right Triangle Given the Hypotenuse and a Leg \(\left({HL}\right)\)

Suppose the hypotenuse \(c\) and leg \(a\) are known for a right triangle.

A right triangle defined by the hypotenuse and a leg.
Figure 1.

To find the other leg \(b,\) we use the Pythagorean theorem:

\[{b^2} = {c^2} - {a^2},\;\; \Rightarrow b = \sqrt {{c^2} - {a^2}} .\]

Since all \(3\) sides are already known, we can determine any angle using a trigonometric function. For example,

\[\sin \alpha = \frac{a}{c}, \;\text{ or }\; \cos \alpha = \frac{b}{c}.\]

The other acute angle can be expressed in terms of the sine or cosine as

\[\cos \beta = \frac{a}{c}, \;\text{ or }\; \sin \beta = \frac{b}{c}.\]

2. Solving a Right Triangle Given Two Legs \(\left({LL}\right)\)

Let a right triangle be defined by legs \(a\) and \(b.\)

A right triangle defined by two legs.
Figure 2.

The hypotenuse \(c\) can be easily found by the Pythagorean theorem:

\[{c^2} = {a^2} + {b^2},\;\; \Rightarrow c = \sqrt {{a^2} + {b^2}} .\]

Now we are at the same point as in the previous case, so the acute angles \(\alpha\) and \(\beta\) can be defined in the same way:

\[\sin \alpha = \frac{a}{c}, \;\text{ or }\; \cos \alpha = \frac{b}{c},\]
\[\cos \beta = \frac{a}{c}, \;\text{ or }\; \sin \beta = \frac{b}{c}.\]

3. Solving a Right Triangle Given the Hypotenuse and an Angle \(\left({HA}\right)\)

A right triangle has a hypotenuse \(c\) and an angle \(\alpha.\)

A right triangle defined by the hypotenuse and an angle.
Figure 3.

We find the legs of the triangle using the Pythagorean trig identity:

\[a = c\sin \alpha ,\;\;b = c\cos \alpha .\]

The other acute angle of the triangle is obviously

\[\beta = {90^\circ} - \alpha .\]

4. Solving a Right Triangle Given a Leg and the Adjacent Angle \(\left({LA_A}\right)\)

Consider a right triangle and suppose its leg \(b\) and the adjacent angle \(\alpha\) are known.

A right triangle defined by a leg and the adjacent angle.
Figure 4.

The hypotenuse can be found by the formula

\[c = \frac{b}{{\cos \alpha }}.\]

The other leg \(a\) is given by

\[a = b\tan \alpha .\]

The angles \(\alpha\) and \(\beta\) are complementary, so

\[\beta = {90^\circ} - \alpha .\]

5. Solving a Right Triangle Given a Leg and the Opposite Angle \(\left({LA_O}\right)\)

The leg \(a\) and the opposite angle \(\alpha\) of a right triangle are known.

A right triangle defined by a leg and the opposite angle.
Figure 5.

In this case, the hypotenuse can be found by the formula

\[c = \frac{a}{{\sin \alpha }}.\]

To determine the other leg \(b,\) we can use the identity

\[b = a\cot \alpha .\]

The angle \(\beta\) is complementary to \(\alpha:\)

\[\beta = {90^\circ} - \alpha .\]

General Case: Solving a Right Triangle Given Two Arbitrary Elements

In general, a right triangle can be defined by two arbitrary distinct elements. One of them must be a metric element (like a leg). For example, we could define a right triangle by its area and perimeter, or say, by an altitude drawn to the hypotenuse and a radius of the inscribed circle. Some pairs of such parameters lead to complex systems of equations. However, there is also a plenty of combinations of this type which are quite solvable. We will consider some of these problems below.

See solved problems on Page 2.

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