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

[eleventhxhour]

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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!

 

[MarkFL]

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.

 

[eleventhxhour]

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

 

[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:

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

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

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

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:

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

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