Given that 𝐴𝐵𝐶 is an isosceles triangle, where the side length 𝐴𝐵 is equal to the side length 𝐴𝐶 is equal to 19 centimeters and the measure of angle 𝐴 is equal to 51 degrees, determine the dot product between the vector 𝐁𝐀 and the vector 𝐁𝐂 correct to the nearest hundredth.
In this question, we’re given some information about triangle 𝐴𝐵𝐶. First, we’re told that this represents an isosceles triangle. Next, we’re told the two equal sides of our isosceles triangle are the sides 𝐴𝐵 and 𝐴𝐶, which are 19 centimeters long. And we’re also told the measure of the angle at 𝐴 is 51 degrees. We need to use this to determine the dot product between vector 𝐁𝐀 and vector 𝐁𝐂. And we need to give our answer to the nearest hundredth.
To answer this question, it’s always a good idea to sketch the information we’re given. So to start, we sketch an isosceles triangle 𝐴𝐵𝐶 with side 𝐴𝐵 and side 𝐴𝐶 equal to 19 centimeters and the angle at 𝐴 51 degrees. Let’s now also sketch our two vectors 𝐁𝐀 and 𝐁𝐂 onto this diagram. 𝐁𝐀 will be the vector which starts at 𝐵 and ends at 𝐴, and 𝐁𝐂 will be the vector which starts at 𝐵 and ends at 𝐶. So now we can see on our sketch the vector 𝐁𝐀 and the vector 𝐁𝐂 which we’re asked to calculate the dot product of.
And we can notice something interesting about these two vectors. We can actually calculate the angle between them. And we actually know a result connecting the dot product of two vectors with the angle between them. So we can use this to evaluate our dot product. We recall if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 must be equal to the dot product between 𝐮 and 𝐯 divided by the modulus of 𝐮 times the modulus of 𝐯. And we know some of this information for the dot product given to us in the question.
For example, we can rewrite this formula with the vectors 𝐁𝐀 and the vectors 𝐁𝐂. We can also include the angle between these vectors on our diagram. We’ll call this 𝜃. We then have the cos of 𝜃 is equal to the dot product between vector 𝐁𝐀 and vector 𝐁𝐂 divided by the modulus of 𝐁𝐀 times the modulus of 𝐁𝐂. So to find the dot product using this method, we’re going to need to find the value of 𝜃 and we’re also going to need to find the modulus of vector 𝐁𝐀 and vector 𝐁𝐂.
We can see one of these straightaway. The modulus of vector 𝐁𝐀 is going to be the length of the side. It’s going to be 19. Next, let’s find the value of 𝜃. To do this, we’re going to need to use the fact that triangle 𝐴𝐵𝐶 is an isosceles triangle. And we know in an isosceles triangle, the angles opposite the equal sides are equal. So our other unknown angle is also equal to 𝜃. But this is not the only thing we know. We also know the sum of the interior angles in a triangle adds to 180 degrees. So we can add these three angles together to give us that 180 degrees should be equal to 𝜃 plus 𝜃 plus 51 degrees.
Now we solve for 𝜃. First, we’ll subtract 51 degrees from both sides of the equation and then simplify. This gives us 129 degrees is equal to two 𝜃. Then we just divide through by two, giving us that 64.5 degrees is equal 𝜃. So the last unknown in our equation is the magnitude of vector 𝐁𝐂. And this is going to be equal to the length of the side of our triangle 𝐵𝐶. Usually, we represent this with a lowercase 𝑎 because it’s opposite the angle capital 𝐴. There’s several different ways we could find this length. For example, we know the opposite angle and the other two sides of our triangle, so we could do this by using the law of cosines.
Applying the law of cosines to this triangle to find our length gives us the modulus of 𝐁𝐂 all squared is equal to 19 squared plus 19 squared minus two times 19 multiplied by 19 times the cos of 51 degrees. And it’s also worth pointing out here we didn’t need to use the law of cosines. We could’ve also used the law of sines. Because we’ve already found the value of 𝜃, we already know all the angles in our triangle. It’s personal preference which one do we use. We’ll just use the law of cosines.
Evaluating the right-hand side of this equation, we get 267.63, and this continues. Then we can find the side length 𝐵𝐶 by taking the square root of both sides. We get the length of the side 𝐵𝐶, or the modulus of vector 𝐁𝐂, is equal to 16.359, and this continues. Now all we need to do is substitute these values into our equation to find the dot product. We get the cos of 64.5 degrees is equal to the dot product between vector 𝐁𝐀 and vector 𝐁𝐂 divided by 19 times 16.359, and this expansion continues. And it’s important we don’t round this value until the very last step in our calculation.
Now to find our dot product, we’re going to need to multiply through by the denominator in our fraction. This gives us the dot product between vector 𝐁𝐀 and vector 𝐁𝐂 is equal to 19 times 16.359, and this continues, multiplied by the cos of 64.5. And if we calculate this expression, we get 133.815, and this expansion continues.
But remember the question wanted us to give our answer to the nearest hundredth. The nearest hundredth means two decimal places. So we need to look at our third decimal place to determine whether we round up or round down. We see the third decimal place is five, which is greater than or equal to five. This means we need to round up. And this gives us our final answer of 133.82. Therefore, we were able to show if 𝐴𝐵𝐶 is an isosceles triangle, where the side 𝐴𝐵 is equal to the side 𝐴𝐶 is equal to 19 centimeters and the measure of angle 𝐴 is 51 degrees, then the dot product between the vectors 𝐁𝐀 and 𝐁𝐂 to the nearest hundredth will be 133.82.