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:

  • The hypotenuse and a leg \(\left({HL}\right),\)
  • Two legs \(\left({LL}\right),\)
  • The hypotenuse and an angle \(\left({HA}\right),\)
  • A leg and the adjacent angle \(\left({LA_A}\right),\)
  • A leg and the opposite angle \(\left({LA_O}\right),\)

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,

\[{\sin \alpha = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\cos \alpha = \frac{b}{c}.}\]

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

\[{\cos \beta = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\sin \beta = \frac{b}{c}.}\]

\(2.\) 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:

\[{\sin \alpha = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\cos \alpha = \frac{b}{c},}\]

\[{\cos \beta = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\sin \beta = \frac{b}{c}.}\]

\(3.\) 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:

\[{a = c\sin \alpha ,\;\;}\kern0pt{b = c\cos \alpha .}\]

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.


Solved Problems

Click or tap a problem to see the solution.

Example 1

A right triangle has a hypotenuse of length \(41\) and a leg of length \(9.\) Find the other leg and two acute angles.

Example 2

The two legs of a right triangle are \(12\) and \(35.\) Find the hypotenuse and two acute angles.

Example 3

In a right triangle, the length of the hypotenuse is \(20\) and an acute angle is \(30^\circ.\) Find the missing elements.

Example 4

In a right triangle, one angle is \(50^\circ\) and the adjacent leg is \(15.\) Find the missing elements.

Example 5

A leg in a right triangle is equal to \(25,\) and the angle opposite to this leg is equal to \(30^\circ.\) Determine the missing elements.

Example 6

Let \(a\) be a leg of a right triangle and \(h\) be the altitude drawn to the hypotenuse. Determine the missing elements of the triangle.

Example 7

The perimeter of a right triangle is \(P\) and its area is \(A.\) Find the length of the hypotenuse.

Example 8

A right triangle with an angle of \(\alpha\) has an area \(A.\) Find the missing elements of the triangle.

Example 9

In a right triangle, the projections of the legs on the hypotenuse are \(m = 9\) and \(n = 16.\) Find the radius of the inscribed circle.

Example 10

The radii of the circumcircle and incircle of a right triangle are \(R\) and \(r.\) Find the legs of the triangle.

Example 1.

A right triangle has a hypotenuse of length \(41\) and a leg of length \(9.\) Find the other leg and two acute angles.

Solution.

According to the notation above, we deal here with an \(HL\) model. Denote \(c = 41\) and \(a = 9.\) The acute angle \(\alpha\) is supposed to be opposite to side \(a.\)

We find the \(b\) using the Pythagorean theorem:

\[{b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{41}^2} – {9^2}} }={ \sqrt {1681 – 81} }={ \sqrt {1600} }={ 40.}\]

The angles \(\alpha\) and \(\beta\) can be specified by the sine function:

\[{\sin \alpha = \frac{a}{c} = \frac{9}{{41}},\;\;}\kern0pt{\sin \beta = \frac{b}{c} = \frac{{40}}{{41}}.}\]

Example 2.

The two legs of a right triangle are \(12\) and \(35.\) Find the hypotenuse and two acute angles.

Solution.

The problem is classified as a \(LL\) model. Let \(a = 12\) and \(b = 35.\) The length of the hypotenuse is given by the Pythagorean theorem:

\[{c = \sqrt {{a^2} + {b^2}} }={ \sqrt {{{12}^2} + {{35}^2}} }={ \sqrt {144 + 1225} }={ \sqrt {1369} }={ 37.}\]

Assuming that the angle \(\alpha\) lies opposite the side \(a,\) we have

\[{\sin \alpha = \frac{a}{c} = \frac{12}{{37}},\;\;}\kern0pt{\sin \beta = \frac{b}{c} = \frac{{35}}{{37}}.}\]

Example 3.

In a right triangle, the length of the hypotenuse is \(20\) and an acute angle is \(30^\circ.\) Find the missing elements.

Solution.

Here we have an \(HA\) scenario. We denote the hypotenuse by \(c\) and the known angle by \(\alpha.\) Using the trigonometric functions, we obtain

\[{a = c\sin \alpha }={ 20\sin {30^\circ} }={ 20 \times \frac{1}{2} }={ 10;}\]

\[{b = c\cos \alpha }={ 20\cos {30^\circ}.}\]

If \(\sin 30^\circ = \large{\frac{1}{2}}\normalsize,\) then

\[{\cos {30^\circ} = \sqrt {1 – {{\sin }^2}{{30}^\circ}} }={ \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} }={ \sqrt {1 – \frac{1}{4}} }={ \sqrt {\frac{3}{4}} }={ \frac{{\sqrt 3 }}{2}.}\]

So the length of the second leg \(b\) is given by

\[{b = 20\cos {30^\circ} }={ 20 \times \frac{{\sqrt 3 }}{2} }={ 10\sqrt 3 .}\]

The complementary angle \(\beta\) is equal to

\[{\beta = {90^\circ} – \alpha }={ {90^\circ} – {30^\circ} }={ {60^\circ}.}\]

Example 4.

In a right triangle, one angle is \(50^\circ\) and the adjacent leg is \(15.\) Find the missing elements.

Solution.

This is a \(LA_A\) model by the above notation. The other acute angle is obviously equal to

\[{\beta = {90^\circ} – \alpha }={ {90^\circ} – {50^\circ} }={ {40^\circ}.}\]

If \(a\) is the leg opposite to the angle \(\alpha,\) we can write

\[{c = \frac{a}{{\sin \alpha }} }={ \frac{{15}}{{\sin {{50}^\circ}}}.}\]

Using a calculator, we find the value of \(\sin 50^\circ:\)

\[\sin {50^\circ} = 0.7660\]

Hence,

\[{c = \frac{{15}}{{\sin {{50}^0}}} }={ \frac{{15}}{{0.7660}} }\approx{ 19.58}\]

The second leg can be found by the Pythagorean theorem:

\[{b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{19.58}^2} – {{15}^2}} }={ \sqrt {158.38} }\approx{ 12.58}\]

Example 5.

A leg in a right triangle is equal to \(25,\) and the angle opposite to this leg is equal to \(30^\circ.\) Determine the missing elements.

Solution.

This triangle matches to the \(LA_O\) definition. Suppose the known leg is \(a\) and the opposite angle is \(\alpha.\) Then, the hypotenuse is given by

\[{c = \frac{a}{{\sin \alpha }} = \frac{{25}}{{\sin {{30}^\circ}}} }={ \frac{{25}}{{\frac{1}{2}}} }={ 50.}\]

There are several ways to find the other leg. Let’s use the Pythagorean theorem:

\[{b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{50}^2} – {{25}^2}} }={ \sqrt {2500 – 625} }={ \sqrt {1875} }={ 25\sqrt 3 .}\]

The complementary angle is

\[{\beta = {90^\circ} – \alpha }={ {90^\circ} – {30^\circ} }={ {60^\circ}.}\]

Example 6.

Let \(a\) be a leg of a right triangle and \(h\) be the altitude drawn to the hypotenuse. Determine the missing elements of the triangle.

Solution.

A right triangle with an altitude h and a leg a.
Figure 6.

For the inner triangle \(BDC,\) the following relationship is valid:

\[{h = a\sin \beta ,}\;\; \Rightarrow {\sin \beta = \frac{h}{a}.}\]

The cosine of \(\beta\) can be found using the Pythagorean trig identity:

\[{\cos \beta = \sqrt {1 – {{\sin }^2}\beta } }={ \sqrt {1 – {{\left( {\frac{h}{a}} \right)}^2}} }={ \sqrt {\frac{{{a^2} – {h^2}}}{{{a^2}}}} }={ \frac{{\sqrt {{a^2} – {h^2}} }}{a}.}\]

Now we can find the hypotenuse \(c:\)

\[{c = \frac{a}{{\cos \beta }} }={ \frac{a}{{\frac{{\sqrt {{a^2} – {h^2}} }}{a}}} }={ \frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }}.}\]

The other leg \(b\) is given by

\[{b = c\sin \beta }={ \frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }} \times \frac{h}{a} }={ \frac{{ah}}{{\sqrt {{a^2} – {h^2}} }}.}\]

Finally, calculate the cosine of \(\alpha:\)

\[{\cos \alpha = \frac{b}{c} }={ \frac{{\frac{{ah}}{{\sqrt {{a^2} – {h^2}} }}}}{{\frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }}}} }={ \frac{h}{a}.}\]

Example 7.

The perimeter of a right triangle is \(P\) and its area is \(A.\) Find the length of the hypotenuse.

Solution.

We denote the legs by \(a, b\) and the hypotenuse by \(c.\) By condition,

\[{a + b + c = P,}\;\; \Rightarrow {a + b = P – c.}\]

The area is given by the formula

\[{A = \frac{{ab}}{2},}\;\; \Rightarrow {2ab = 4A.}\]

Consider the sum of the legs squared:

\[{{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} }={ {c^2} + 4A,}\]

where we used the Pythagorean theorem: \({a^2} + {b^2} = {c^2}.\)

From the other side, we have

\[a + b = P – c.\]

Therefore, we get the following equation for \(c:\)

\[{\left( {P – c} \right)^2} = {c^2} + 4A.\]

It has a simple solution:

\[\require{cancel}{{P^2} – 2Pc + \cancel{c^2} = \cancel{c^2} + 4A,}\;\; \Rightarrow {2Pc = {P^2} – 4A,}\;\; \Rightarrow {c = \frac{{{P^2} – 4A}}{{2P}}.}\]

Example 8.

A right triangle with an angle of \(\alpha\) has an area \(A.\) Find the missing elements of the triangle.

Solution.

The area of a right triangle is given by the formula

\[A = \frac{{ab}}{2},\]

where \(a\) and \(b\) are the legs of the triangle.

Let \(c\) be the (unknown) hypotenuse of the triangle. The legs can be expressed in terms of \(c\) and \(\alpha:\)

\[{a = c\sin \alpha ,\;\;}\kern0pt{b = c\cos \alpha .}\]

We assume here that the angle \(\alpha\) is opposite to the side \(a.\)

Then we have

\[{A = \frac{{ab}}{2} }={ \frac{1}{2}{c^2}\sin \alpha \cos \alpha .}\]

Solve this equation for \(c:\)

\[{{c^2} = \frac{{2A}}{{\sin \alpha \cos \alpha }},}\;\; \Rightarrow {c = \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} .}\]

We now derive the expressions for legs:

\[\require{cancel}{a = c\sin \alpha }={ \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} \times \sin \alpha }={ \sqrt {\frac{{2A{{\sin }^{\cancel{2}}}\alpha }}{{\cancel{\sin \alpha} \cos \alpha }}} }={ \sqrt {\frac{{2A\sin \alpha }}{{\cos \alpha }}} }={ \sqrt {2A\tan \alpha } ;}\]

\[{b = c\cos \alpha }={ \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} \times \cos \alpha }={ \sqrt {\frac{{2A{{\cos }^{\cancel{2}}}\alpha }}{{\sin \alpha \cancel{\cos \alpha }}}} }={ \sqrt {\frac{{2A\cos \alpha }}{{\sin \alpha }}} }={ \sqrt {2A\cot \alpha } .}\]

The complentary angle is obviously equal to

\[\beta = \frac{\pi }{2} – \alpha .\]

Example 9.

In a right triangle, the projections of the legs on the hypotenuse are \(m = 9\) and \(n = 16.\) Find the radius of the inscribed circle.

Solution.

A right triangle with given projections of legs to the hypotenuse.
Figure 8.

Applying the Pythagorean theorem to the triangles \(ADC\) and \(BDC,\) we get

\[{{b^2} – {m^2} = {b^2} – {9^2} = {h^2},\;\;}\kern0pt{{a^2} – {n^2} = {a^2} – {16^2} = {h^2}.}\]

Hence,

\[{{b^2} – 81 = {a^2} – 256, \;\;}\kern0pt{\text{or }\; {b^2} = {a^2} – 175.}\]

The second equation is given by Pythagorean theorem for the triangle \(ABC:\)

\[{{a^2} + {b^2} = {c^2} }={ {\left( {m + n} \right)^2} }={ {25^2} }={ 625.}\]

Thus, we have a system of two equations with the variables \(a^2\) and \(b^2:\)

\[\left\{ \begin{array}{l} {b^2} = {a^2} – 175\\ {a^2} + {b^2} = 625 \end{array} \right..\]

Solving the system yields the following results:

\[{{a^2} + {a^2} – 175 = 625,}\;\; \Rightarrow {2{a^2} = 800,}\;\; \Rightarrow {{a^2} = 400,}\;\; \Rightarrow {a = 20;}\]

\[{{b^2} = {a^2} – 175 }={ 400 – 175 }={ 225,}\;\; \Rightarrow {b = 15.}\]

The radius of the inscribed circle is given by

\[{r = \frac{{a + b – c}}{2} }={ \frac{{20 + 15 – 25}}{2} }={ 5.}\]

Example 10.

The radii of the circumcircle and incircle of a right triangle are \(R\) and \(r.\) Find the legs of the triangle.

Solution.

Let \(AB = c\) is the hypotenuse and \(BC = a,\) \(AC = b\) are the legs. The figure \(CKOL\) is a square with side \(r.\) Suppose that the length of the line segment \(AK\) is \(x.\) The secant segments \(AK\) and \(AM\) have the same length \(x.\)

A right triangle defined by the radii of the circumcircle and incircle.
Figure 7.

Recall that the hypotenuse length is twice the radius of the circumcircle: \(c = 2R.\) The secant segments \(BM\) and \(BL\) have the same length equal to \(2R – x.\)

We express the sides of the triangle in terms of \(x\) and the radii \(R\) and \(r:\)

\[{a = AC = r + x,\;\;}\kern0pt{b = BC = r + 2R – x,\;\;}\kern0pt{c = AB = 2R.}\]

By the Pythagorean theorem,

\[{{a^2} + {b^2} = {c^2},}\;\; \Rightarrow {{\left( {r + x} \right)^2} + {\left( {r + 2R – x} \right)^2} = 4{R^2}.}\]

Now we need to solve the equation for \(x.\) To simplify it, we use the identities for the square of a sum and the square of a trinomial:

\[{\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2},\]

\[{{\left( {x + y + z} \right)^2} }={ {x^2} + {y^2} + {z^2} }+{ 2xy + 2xz + 2yz.}\]

Hence,

\[{{\left( {r + x} \right)^2} + {\left( {r + 2R – x} \right)^2} }={ 4{R^2},}\]

\[ \Rightarrow {{r^2} + \cancel{2rx} + {x^2} }+{ {r^2} + \cancel{4{R^2}} + {x^2} }+{ 4rR – \cancel{2rx} – 4Rx }={ \cancel{4{R^2}},}\]

\[ \Rightarrow 2{x^2} – 4Rx + 2{r^2} + 4rR = 0,\]

\[ \Rightarrow {x^2} – 2Rx + {r^2} + 2rR = 0.\]

This quadratic equation has two solutions:

\[{{x_{1,2}} }={ \frac{{2R \pm \sqrt {4{R^2} – 4{r^2} – 8rR} }}{2} }={ R \pm \sqrt {{R^2} – {r^2} – 2rR} .}\]

If we take the first root

\[{x_1} = R – \sqrt {{R^2} – {r^2} – 2rR},\]

we get the following lengths of the legs:

\[{a = r + x }={ r + R – \sqrt {{R^2} – {r^2} – 2rR} ,}\]

\[{b = r + 2R – x }={ r + 2R – R }+{ \sqrt {{R^2} – {r^2} – 2rR} }={ r + R + \sqrt {{R^2} – {r^2} – 2rR} .}\]

It is clear that the second root \(x_2\) produces the same legs only rearranged in places. Thus, the final answer is

\[{a = r + R – \sqrt {{R^2} – {r^2} – 2rR} ,}\]

\[b = r + R + \sqrt {{R^2} – {r^2} – 2rR} .\]

Let’s substitute some sample values. If \(R = 10\) and \(r = 4,\) we obtain

\[{a = 4 + 10 }-{ \sqrt {{{10}^2} – {4^2} – 2 \cdot 4 \cdot 10} }={ 4 + 10 + \sqrt 4 }={ 16,}\]

\[b = 4 + 10 – \sqrt 4 = 12.\]