Where Did I Go Wrong in Calculating the Value of x in These Triangle Problems?

[eleventhxhour]

View attachment 2480

3) Determine the value of x to the nearest centimetre.

So for 3a) I found the angles of each of the sides by subtracting 180 (D is 55 on the first triangle, and 45 degrees on the second). Then I found "f" (the side opposite angle F) by using the Sine law, and I got 18cm. Then I used the sine law to find the length of the dashed line and I got 8.6cm. Then with this, I used the sine law again to calculate x. In the end, I got 12.2 cm. However, the answer in the textbook is 15cm. What did I do wrong?

For 3b, I did something similar. I found that the smaller triangle is isosceles, so the angles are: D is 70 degrees, B is 40 degrees and the one that isn't labelled is also 70 degrees. On the bigger triangle, angle B is 63 degrees. Then I used the sine law to calculate the horizontal line from B. I got 21.9cm. Then I used the sine law again to calculate x, which I found to be 48.2. However, the textbook says that x = 37.9cm. What did I do wrong?

Thanks!

 

[MarkFL]

For 3a) I get:

\(\displaystyle x=\frac{15\sqrt{2}\sin\left(35^{\circ}\right)}{\sin\left(55^{\circ}\right)}\text{ cm}\approx15\text{ cm}\)

For 3b) I get:

\(\displaystyle x=\frac{15\sin\left(70^{\circ}\right)}{\sin\left(27^{\circ}\right)\sin\left(55^{\circ}\right)}\text{ cm}\approx38\text{cm}\)

Your textbook is correct. I am unsure what you've done...you say you applied the sine law, but how you applied it I am not sure.

 

[eleventhxhour]

For 3a) I get:

\(\displaystyle x=\frac{15\sqrt{2}\sin\left(35^{\circ}\right)}{\sin\left(55^{\circ}\right)}\text{ cm}\approx15\text{ cm}\)

For 3b) I get:

\(\displaystyle x=\frac{15\sin\left(70^{\circ}\right)}{\sin\left(27^{\circ}\right)\sin\left(55^{\circ}\right)}\text{ cm}\approx38\text{cm}\)

Your textbook is correct. I am unsure what you've done...you say you applied the sine law, but how you applied it I am not sure.

Alright, so I did 3a) again and got the correct answer. But I'm still getting around 48cm for 3b). Here's what I did:

View attachment 2481

 

[MarkFL]

You can only apply the Pythagorean theorem to right triangles. The isosceles triangle has a $70^{\circ}$, and therefore the other two equal angles $\theta$ are:

\(\displaystyle 70^{\circ}+2\theta=180^{\circ}\)

\(\displaystyle 2\theta=110^{\circ}\)

\(\displaystyle \theta=55^{\circ}\)

And so, using the Law of Sines, we find:

\(\displaystyle \frac{\overline{BC}}{\sin\left(70^{\circ}\right)}=\frac{15}{\sin\left(55^{\circ}\right)}\)

Hence:

\(\displaystyle \overline{BC}=\frac{15\sin\left(70^{\circ}\right)}{\sin\left(55^{\circ}\right)}\)

Now, using the definition of the sine function, we find:

\(\displaystyle \sin\left(27^{\circ}\right)=\frac{\overline{BC}}{x}\)

\(\displaystyle x=\frac{\overline{BC}}{\sin\left(27^{\circ}\right)}=\frac{15\sin\left(70^{\circ}\right)}{\sin\left(27^{\circ}\right)\sin\left(55^{\circ}\right)}\)

 

[CellCoree]

Hello,

Thank you for sharing your work and asking for feedback. It seems like you have a good understanding of the concepts and equations used to solve these problems. However, there may be some small errors in your calculations that resulted in the incorrect answers.

For 3a), it would be helpful to see your work to determine where the discrepancy may have occurred. One possibility is that you may have used the incorrect angle for the side "f" when applying the sine law. Double-checking your calculations step by step can help identify any errors.

For 3b), it looks like you may have mixed up the angles in the smaller triangle. The angles should be D=70 degrees, B=70 degrees, and the one not labeled = 40 degrees. This may have affected your calculation for the horizontal line from B, leading to the incorrect value for x.

Overall, it's important to carefully check your work and make sure you are using the correct values for each step. It's also helpful to double-check your answer using a different method or equation to ensure accuracy. Keep up the good work and don't be discouraged by small errors, as they are a natural part of the learning process.

Best of luck with your future scientific endeavors!