3.3: Solving Right Triangles

Example \(\PageIndex{1}\)

Find the side length opposite the \(50^{\circ}\) angle in Figure \(\PageIndex{3}\).

Solution

In this triangle, the trigonometric function that relates the missing side to the given information is the sine:\[\sin\left(50^{\circ}\right)=\dfrac{\text{opposite}}{\text{hypotenuse}}\nonumber\]We use a calculator to find an approximate value for the sine of \(50^{\circ}\), filling in the lengths of the hypotenuse and the opposite side to get\[0.7660 \approx \dfrac{x}{18}.\nonumber\]We solve for \(x\) to find\[x \approx 18(0.7660) = 13.788.\nonumber\]To two decimal places, the length of the opposite side is approximately \(13.79\) centimeters.

Example \(\PageIndex{2}\)

In the right triangle \(\triangle ABC\), \(\angle A = 37^{\circ}\) and (non-hypotenuse) side \(b = 11\). Solve the triangle.

Solution

Since \(\angle A = 37^{\circ}\), side \(a\) is not the hypotenuse of the right triangle. Moreover, we were told that side \(b\) is not the hypotenuse. Thus, \(c\) must be the hypotenuse and \(\angle C = 90^{\circ}\). Creating a rough sketch of \(\triangle ABC\), we get

Using the Triangle Sum, we quickly compute \(\angle B = 90^{\circ}-37^{\circ}=53^{\circ}\). We now concentrate on finding the missing sides by using the trigonometric functions and the given information. The tangent relates \(37^{\circ}\) to sides \(a\) and \(b\). Therefore,\[\tan\left(37^{\circ}\right) = \dfrac{a}{11} \implies a = 11\tan\left(37^{\circ}\right) \approx 8.2891.\nonumber\]Finally, we use the cosine to relate \(37^{\circ}\) to the given side \(b\) and the hypotenuse \(c\).\[\cos\left(37^{\circ}\right) = \dfrac{11}{c} \implies c = \dfrac{11}{\cos\left(37^{\circ}\right)} \approx 13.7735.\nonumber\]

Example \(\PageIndex{3}\)

Solve \(\triangle ABC\) given that \(a = 3\) and \(c = 5\).

Solution

It's always best to start by drawing a triangle.

We can use the Pythagorean Theorem to compute the length of side \(b\), or we could recognize that this is a Pythagorean Triple. No matter how we compute it, \(b = 4\). Now, we need to find the missing angles. Luckily, we were given that this is a right triangle, so once we find either \(\angle A\) or \(\angle B\), we can use the Triangle Sum to find the other angle. Let's go ahead and find \(\angle A\). Since \(\sin\left(\angle A\right) = \frac{a}{c} = \frac{3}{5}\), we use a calculator to compute \(\angle A = \sin^{-1}\left(\frac{3}{5}\right) \approx 36.8699^{\circ}\). Thus,\[\angle B = 180^{\circ}-\angle C-\angle A \approx 180^{\circ}-90^{\circ}-36.8699^{\circ} \approx 53.1301^{\circ}.\nonumber\]It's often best to organize all of this information in a table like so:\[\begin{array}{rclcrcl} \angle A & \approx & 36.8699^{\circ} & \quad & a & = & 3 \\[6pt] \angle B & \approx & 53.1301^{\circ} & \quad & b & = & 4 \\[6pt] \angle C & = & 90^{\circ} & \quad & c & = & 5 \\[6pt] \end{array}\nonumber\]

Example \(\PageIndex{4}\)

A 10-foot ladder is leaning against a wall. If the ladder's base is too far from the wall, the base can slide out. If the ladder base is too close to the wall, there’s a risk that the ladder could tip over backward. The leaning ladder should make a \(75^{\circ}\) angle with the ground for safety reasons.

1. How far should the base of the ladder be from the wall?
2. How far up the wall will the top of the ladder reach?

Solutions

1. The distance between the ladder's base and the wall is the side adjacent to the \(75^{\circ}\) angle. We can find the side adjacent to the \(75^{\circ}\) angle using the cosine ratio.\[\begin{array}{rrcl} & \cos\left(75^{\circ}\right) & = & \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[6pt] \implies & 0.2588 & \approx & \dfrac{b}{10} \\[6pt] \implies & 10(0.2588) & \approx & b \\[6pt] \implies & b & \approx & 2.588 \\[6pt] \end{array}\nonumber\]The ladder's base should be about 2.6 feet from the wall.
2. We could use the Pythagorean Theorem to find how far up the wall the ladder will reach. However, using the given information rather than the calculated values to find the other unknown parts is better. We will use the sine ratio.\[\begin{array}{rrcl} & \sin\left(75^{\circ}\right) & = & \dfrac{\text{opposite}}{\text{hypotenuse}} \\[6pt] \implies & 0.9659 & \approx & \dfrac{a}{10} \\[6pt] \implies & 10(0.9659) & \approx & a \\[6pt] \implies & a & \approx & 9.659 \\[6pt] \end{array}\nonumber\]The ladder will reach about 9.7 feet.

Example \(\PageIndex{5}\)

The vertex angle of an isosceles triangle is \(34^{\circ}\), and the equal sides are 16 meters long. Find the altitude of the triangle.

Solution

The triangle described is not a right triangle. However, the altitude of an isosceles triangle bisects the vertex angle. It divides the triangle into two congruent right triangles, as shown in Figure \(\PageIndex{9}\). The 16-meter side becomes the hypotenuse of the right triangle. The altitude, \(h\), of the original triangle is the side adjacent to the \(17^{\circ}\) angle.

Which of the three trigonometric ratios is helpful in this problem? The cosine is the ratio that relates the hypotenuse and the adjacent side. Therefore, we will begin with the equation:\[\cos\left(17^{\circ}\right)=\dfrac{\text{adjacent}}{\text{hypotenuse}}\nonumber\]We use a calculator to find \(\cos\left(17^{\circ}\right)\) and fill in the lengths of the sides to get\[0.9563 \approx \dfrac{h}{16}.\nonumber\]Solving for \(h\) gives\[h \approx 16(0.9563) = 15.3008.\nonumber\]The altitude of the triangle is about \(15.3\) meters long.

Example \(\PageIndex{6}\)

Figure \(\PageIndex{10}\) shows a circle of radius \(r\) and a right triangle with side \(a\) along the radius of the circle and side \(b\) tangent to the circle.

If \(\angle B = 50^{\circ}\) and \(b = 10\) inches, find the radius of the circle and the partial length of the hypotenuse, \(x\), that sits outside the circle. Round your answers to the nearest tenth.

Solution

Throughout Trigonometry, you will be faced with thought-provoking questions. The beautiful thing about right triangle trigonometry is that you typically need to find a trigonometric function that relates the given information to the missing information. In this case, we are given \(\angle B\) and side \(b\). We are initially asked for the radius of the circle, which is the same as the side length \(a\) of the right triangle. The trigonometric function relating \(\angle B\) to sides \(a\) and \(b\) is the tangent:\[\tan\left(\angle B\right) = \dfrac{b}{a} = \dfrac{b}{r}.\nonumber\]Therefore,\[\tan\left(50^{\circ}\right) = \dfrac{10}{r} \implies r = \dfrac{10}{\tan\left(50^{\circ}\right)}.\nonumber\]Hence, the radius of the circle is \(\frac{10}{\tan\left(50^{\circ}\right)} \approx 8.4\) inches.

Now we need to find the length \(x\). Again, we want to find a trigonometric function that relates the information to what we seek. Unfortunately, we are looking for only part of the hypotenuse of the right triangle; however, the entire hypotenuse is \(x+r\), and we just found \(r\). The sine is the trigonometric function that relates \(\angle B\) to side \(b\) and the hypotenuse:\[\sin\left(\angle B\right) = \dfrac{b}{x+r} \implies x+r = \dfrac{b}{\sin\left(\angle B\right)} \implies x = \dfrac{b}{\sin\left(\angle B\right)}-r.\nonumber\]Substituting the given values for \(\angle B\) and side \(b\), along with the exact value for \(r\), we get\[x = \dfrac{10}{\sin\left(50^{\circ}\right)}-\dfrac{10}{\tan\left(50^{\circ}\right)} \approx 4.66\text{ inches}.\nonumber\]

Example \(\PageIndex{7}\)

In Figure \(\PageIndex{11}\), \(x = 5\), \(\angle A = 25^{\circ}\), and \(\angle BDC = 48^{\circ}\). Find \(y\).

Solution

It is helpful to redraw only \(\triangle BCD\) and then \(\triangle ABC\), noticing that they share the common side \(h\). For \(\triangle BCD\), we only have one piece of given information, \(\angle BDC\); however, we need to find \(y\), and it will be useful to work with \(h\) since both triangles share that side (I call this the "link" between the triangles). The trigonometric function relating these is the tangent:\[\tan\left(\angle BDC\right) = \dfrac{h}{y} \implies y\tan\left(\angle BDC\right) = h.\nonumber\]Hence,\[h = y\tan\left(48^{\circ}\right).\nonumber\]Moving to \(\triangle ABC\), since we were given \(x\) and \(\angle A\), it seems best to use the tangent again to get \(h\) involved. Therefore,\[\tan\left(\angle A\right) = \dfrac{h}{x+y} \implies (x+y)\tan\left(\angle A\right) = h.\nonumber\]That is,\[h = (5+y)\tan\left(25^{\circ}\right).\nonumber\]We now equate these two versions of \(h\) to each other (since they both equal \(h\), they are both equal to each other).\[\begin{array}{rcl} y\tan\left(48^{\circ}\right) = (5+y)\tan\left(25^{\circ}\right) & \implies & y\tan\left(48^{\circ}\right) = 5\tan\left(25^{\circ}\right)+y\tan\left(25^{\circ}\right) \\[6pt] & \implies & y\tan\left(48^{\circ}\right)-y\tan\left(25^{\circ}\right) = 5\tan\left(25^{\circ}\right) \\[6pt] & \implies & y\left(\tan\left(48^{\circ}\right)-\tan\left(25^{\circ}\right)\right) = 5\tan\left(25^{\circ}\right) \\[6pt] & \implies & y = \dfrac{5\tan\left(25^{\circ}\right)}{\tan\left(48^{\circ}\right)-\tan\left(25^{\circ}\right)} \\[6pt] & \implies & y \approx 3.62 \\[6pt] \end{array}\nonumber\]