# Precalculus

## Trigonometry # Solving Right Triangles

## 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:

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

### 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.

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.

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 $$x.$$

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

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