Precalculus

Trigonometry

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Solving Right Triangles

Solved Problems

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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:

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

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

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

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 \(\sin 30^\circ = \frac{1}{2},\) 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

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

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 \(\alpha:\)

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

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:

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

The complementary 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 7.

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

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

Hence,

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

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 8.

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 \(r:\)

\[a = AC = r + x,\;\;b = BC = r + 2R - x,\;\;c = AB = 2R.\]

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

\[a = 4 + 10 - \sqrt {{{10}^2} - {4^2} - 2 \cdot 4 \cdot 10} = 4 + 10 + \sqrt 4 = 16,\]
\[b = 4 + 10 - \sqrt 4 = 12.\]
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