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

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In the discussion, the user attempts to calculate the value of x in two triangle problems using the sine law but arrives at incorrect answers compared to the textbook. For the first triangle (3a), the user initially calculated x as 12.2 cm, while the correct answer is 15 cm. In the second triangle (3b), the user found x to be 48.2 cm, but the textbook states it should be 37.9 cm. After reevaluating the calculations, the user realizes the correct application of the sine law leads to the expected results. The final calculations confirm the textbook answers, emphasizing the importance of correctly applying trigonometric principles.
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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!
 

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For 3a) I get:

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

For 3b) I get:

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

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.
 
MarkFL said:
For 3a) I get:

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

For 3b) I get:

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

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
 

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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:

$$70^{\circ}+2\theta=180^{\circ}$$

$$2\theta=110^{\circ}$$

$$\theta=55^{\circ}$$

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

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

Hence:

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

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

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

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