Finding Trigonometric Ratios

Finding Trigonometric Ratios exercise

• Determine the trigonometric ratios of the pyramid.

  Hints

  The Pharoah's name, SOHCAHTOA, is a mnemonic device that helps us remember the trig ratios for $sin$, $cos$, and $tan$. It breaks into three parts, one for each trig ratio:

  SOH, CAH, and TOA.

  S represents sin, C represents cos, and T represents tan. O represents the side length Opposite to the angle, A represents the side length Adjacent to the angle, and H represents the hypotenuse.

  The trigonometric ratios, or trig ratios, can be applied to right triangles.

  The mnemonic device for each trig ratio starts with the first letter of the name of that ratio. For example, TOA starts with T, which stands for tan.

  Next is the first letter of the side whose length is in the numerator of the trig ratio. Last is the first letter of the side whose length is in the denominator of the trig ratio. For example TOA means that the Opposite side is in the numerator of this trig ratio, and the Adjacent side is in the denominator.

  Solution

  In order to determine the trig ratios, we need a right triangle. The pyramid can be cut in half by drawing a line from the peak to the center of the base. This gives us two right triangles. We can use one of these right triangles to determine trig ratios.

  The mnemonic device for each trig ratio starts with the first letter of the name of that ratio. For example, SOH starts with S, which stands for $sin$.

  Next is the first letter of the side whose length is in the numerator of the trig ratio. Last is the first letter of the side whose length is in the denominator of the trig ratio. For example SOH means that the Opposite side is in the numerator of this trig ratio, and the Hypotenuse is in the denominator. Or in other words:

  $~$

  $sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$

  $~$

  Similarly, CAH means that:

  $cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

  $~$

  And TOA means that:

  $tan = \dfrac{\text{opposite}}{\text{adjacent}}$

  $~$

  Let's identify the three side lengths that we need to calculate these ratios.

  The opposite side to angle $A$ is $479.16$ ft long. This is the side that does not have the angle $A$ at either of its end points.

  The adjacent side is the side between angle $A$ and the right angle. So its length is $377.50$ ft.

  The hypotenuse is always the side that is opposite to the right angle. In this case it is $610$ ft long.

  Now we can use the equations for the trig ratios, and the required side lengths, to calculate each ratio:

  $~$

  $sin = \dfrac{\text{opposite}}{\text{hypotenuse}}= \dfrac{479.16}{610}$

  So we have that $sin = 0.786$.

  $~$

  $cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}= \dfrac{377.50}{610}$

  So we have that $cos = 0.619$.

  $~$

  $tan = \dfrac{\text{opposite}}{\text{adjacent}}= \dfrac{479.16}{377.50}$

  So we have that $tan = 1.27$.

• Find the missing length.

  Hints

  Remember the mnemonic device "TOA".

  The tangent ratio for an angle is equal to the length of the opposite side over the length of the adjacent side.

  Once you calculate the length of the adjacent side using the tangent ratio, you will need to double it to get the full length of the base of the pyramid.

  Solution

  The Pharaoh knows that for angle $A$, he has

  $tan = 1.269$.

  The height of the pyramid is opposite side to the angle $A$, and measures $20$ ft.

  The mnemonic $TOA$ reminds him that $tan$ is equal to the opposite side over the adjacent side length:

  $tan = \dfrac{\text{opposite}}{\text{adjacent}}$.

  He can put in the known values and solve for the missing value:

  $\begin{array}{rcl} 1.269 & = & \dfrac{20}{\text{adjacent}}\\ \color{#669900}{\times \, \text{adjacent} \, \div \, 1.269} & &\color{#669900}{\times \, \text{adjacent} \, \div \, 1.269}\\ \text{adjacent} & = & \dfrac{20}{1.269}\\ \text{adjacent} & = & 15.76 \end{array}$

  He knows that the adjacent side length to angle $A$ in the right triangle is only half the length of the base of the pyramid. So he multiplies this adjacent side length by two to get the length of the base of the pyramid, $31.52$ ft.

• Highlight the Pharaoh's mistakes.

  Hints

  Don't assume that anything is correct, even the first equation.

  When the Pharaoh is distracted, he sometimes flips the numerator and denominator in his equations.

  The base is equal to two times the length of the side adjacent to angle $A$.

  The Pharaoh should use the tangent ratio to start his calculations.

  Solution

  To start with, we know the correct values of the three trig ratios for angle $A$, and we know the height of the desired pyramid. We want to find the length of one side of the base of the pyramid.

  The Pharaoh's mistakes are noted below with an '✗' next to them. The rest is the correct solution to this problem.

  We can find the base, $b$, as being equal to two times the adjacent side to angle $A$:

  $b = 2 \times adj$.

  The height of the pyramid is the opposite side to angle $A$. We have the length of the opposite side, and we need the length of the adjacent side. So we know that we need a trig ratio that includes both the opposite and adjacent sides. The tangent ratio meets this requirement:

  $tan = \dfrac{\text{opp}}{\text{adj}}$.

  ✗ The Pharaoh wrote sin instead of $tan$.

  We know that $tan A = 1.269$, and that the opposite side length is $12$. So:

  $1.269 = \dfrac{12}{\text{adj}}$.

  Using opposite operations:

  $\begin{array}{rcl} 1.269 & = & \dfrac{12}{\text{adj}}\\ \color{#669900}{\div \, 1.269 \, \times \, \text{adj}} & &\color{#669900}{\div \, 1.269 \, \times \, \text{adj}}\\ \text{adj} & = & \dfrac{12}{1.269}\\ \text{adj} & = & 9.5 \end{array}$

  ✗ The Pharaoh switched the numerator and the denominator.

  We know that $b = 2 \times adj$. Now we know the length of the adjacent side, so we can find the side length of the base $b$:

  $b = 2 \times 9.5$, so

  $b = 19$.

  ✗ The Pharaoh multiplied the adjacent side by 3, instead of two. As a result, he got the wrong side length for the base, even though he found the correct length of the side adjacent to angle $A$.

• Determine the appropriate trig ratio.

  Hints

  The hypotenuse is always the side that is opposite the right angle. The opposite and adjacent sides are always relative to the target angle. Here, the hypotenuse is $5$ units long. The opposite side is $4$ units long, and the adjacent side is not labeled with a length.

  Each triangle has two known side lengths. The sine, cosine, or tangent ratio will include both of the known side lengths.

  Solution

  For each triangle we have a target angle and we know two side lengths. We must identify a trig ratio that can be found for the target angle using the known side lengths.

  Let's start by identifying which side lengths we know, with respect to the target angle.

  For the first pendant, the side with a length of $4$ units is opposite angle $b$. We can verify this by observing that angle $b$ is not at either end point of this side. Similarly, the side with a length of $5$ is the hypotenuse because it is opposite the right angle.

  Sin is the trig ratio that includes the opposite side, and the hypotenuse. So we should use sin to characterize angle $b$ in the first pendant.

  For the second pendant, we have an hourglass shape. The top half of the hourglass is a right triangle. Angle $b$ is the target angle. The side with a length of $2$ units is the adjacent side to angle $b$, because is goes between angle $b$ and the right angle. The side with a length of $3$ units is the hypotenuse because it is opposite the right angle. The cos trig ratio includes both the adjacent side and the hypotenuse, so we should use this ratio for angle $b$ in this triangle.

  For the third pendant, the side with a length of $4$ units is adjacent side because it is between angle $b$ and the right angle. The side with a length of two is opposite to angle $b$; $tan$ is the trig ratio that includes both the opposite and the adjacent sides.

  For the fourth pendant, we have a rectangle that is split to form two right triangles. Looking at angle a in the upper triangle, the side with a length of $3$ units is the adjacent side because it is between angle $b$ and the right angle. The side with a length of $4$ units is the hypotenuse because it is opposite the right angle.

• Identify the sides.

  Hints

  Remember that the Pharaoh's name, SOHCAHTOA, can help you remember the trig ratios.

  The Pharaoh's name breaks into three syllables, each of which helps you remember one of the trig ratios. The three letters stand for the trig ratio, and the sides that belong in the numerator and denominator of that ratio.

  Solution

  The mnemonic device SOHCAHTOA breaks into three parts, one for each of the trig ratios. They are SOH, CAH, and TOA. Each starts with the first letter of the name of that ratio. For example, SOH starts with S, which stands for $sin$.

  The next letter stands for the side whose length is in the numerator of the trig ratio. The last letter stands for the side whose length is in the denominator of the trig ratio. For example SOH means that the opposite side is in the numerator of this trig ratio, and that the hypotenuse is in the denominator. In other words:

  $~$

  $sin = \dfrac{\text{opposite}}{\text{adjacent}}$

  $~$

  Similarly, CAH means that:

  $cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

  $~$

  And TOA means that:

  $tan = \dfrac{\text{adjacent}}{\text{opppsite}}$

  $~$

  The opposite side to angle $A$ is the side that does not have the angle $A$ at either of its end points.

  The adjacent side is the side between angle $A$ and the right angle.

  The hypotenuse is always the side that is opposite to the right angle.

• Complete the trig ratios.

  Hints

  Remember that the opposite and adjacent sides are named with respect to the target angle. So the opposite side for angle $B$ is different than the opposite side for angle $A$.

  It may help to write out the equations in numbers and symbols, in order to understand how to fill in the blanks in the sentences.

  Remember, the Pharaoh's name, SOH-CAH-TOA, can help you remember what sides are in the numerator and the denominator of each trig ratio.

  Solution

  Recall the trig ratios:

  $sin = \dfrac{\text{opp}}{\text{hyp}}$

  $cos = \dfrac{\text{adj}}{\text{hyp}}$

  $tan = \dfrac{\text{opp}}{\text{adj}}$

  For the first triangle, the hypotenuse is 206 ft long, and does not change depending on the angle. For angle $A$, the opposite side is 100 ft long because this is the side that does not have angle $A$ at either of its end points. The adjacent side is 180 ft long because this is the side that goes between angle $A$ and the right angle. For angle $B$, the opposite side is 180 ft long, and the adjacent side is 100 ft long.

  So we can find cosine ratio as:

  $cos \, A = \dfrac{180}{206}$

  $cos \, A = 0.874$

  So the cosine of angle A is equal to 180 divided by 206. The ratio is 0.874.

  Similarly:

  $tan \, B = \dfrac{180}{100}$

  $tan \, B = 1.8$

  $sin \, A = \dfrac{100}{206}$

  $sin \, A = 0.485$

  And for the second triangle:

  $sin \, A = \dfrac{40}{50}$

  $sin \, A = 0.8$

  $tan \, B = \dfrac{30}{40}$

  $tan \, B = 0.75$