Let's solve questions 6, 7, and 8 from your second image using the same method shown in Example 6, which involves solving right triangles.

Question 6:

For triangle $\triangle ABC$ with:

• $\angle C = 90^\circ$
• $\angle B = 66^\circ$
• $AC = 5.5 \, \text{m}$ (adjacent to angle $B$)

(a) Find the measure of side $a$ (opposite to $\angle B$).

Using the trigonometric function for opposite and adjacent sides: $\tan(B) = \frac{a}{AC}$ $\tan(66^\circ) = \frac{a}{5.5}$ $a = 5.5 \times \tan(66^\circ) \approx 5.5 \times 2.246 = 12.35 \, \text{m}$

(b) Find the measure of side $b$ (hypotenuse).

Using cosine for adjacent and hypotenuse: $\cos(B) = \frac{AC}{b}$ $\cos(66^\circ) = \frac{5.5}{b}$ $b = \frac{5.5}{\cos(66^\circ)} \approx \frac{5.5}{0.4067} = 13.52 \, \text{m}$

Question 7:

For triangle $\triangle ABC$ with:

• $\angle C = 90^\circ$
• $\angle B = 70^\circ$
• $AB = 35.5 \, \text{cm}$ (hypotenuse)

(a) Find the measure of side $a$ (opposite to $\angle B$).

Using sine for opposite and hypotenuse: $\sin(B) = \frac{a}{AB}$ $\sin(70^\circ) = \frac{a}{35.5}$ $a = 35.5 \times \sin(70^\circ) \approx 35.5 \times 0.9397 = 33.35 \, \text{cm}$

(b) Find the measure of side $b$ (adjacent to $\angle B$).

Using cosine for adjacent and hypotenuse: $\cos(B) = \frac{b}{AB}$ $\cos(70^\circ) = \frac{b}{35.5}$ $b = 35.5 \times \cos(70^\circ) \approx 35.5 \times 0.342 = 12.15 \, \text{cm}$

Question 8:

For triangle $\triangle ABC$ with:

• $\angle C = 90^\circ$
• $\angle B = 63^\circ$
• $AB = 100 \, \text{m}$ (hypotenuse)

(a) Find the measure of side $a$ (opposite to $\angle B$).

Using sine for opposite and hypotenuse: $\sin(B) = \frac{a}{AB}$ $\sin(63^\circ) = \frac{a}{100}$ $a = 100 \times \sin(63^\circ) \approx 100 \times 0.8910 = 89.10 \, \text{m}$

(b) Find the measure of side $b$ (adjacent to $\angle B$).

Using cosine for adjacent and hypotenuse: $\cos(B) = \frac{b}{AB}$ $\cos(63^\circ) = \frac{b}{100}$ $b = 100 \times \cos(63^\circ) \approx 100 \times 0.453 = 45.30 \, \text{m}$

Would you like a more detailed explanation for any of these steps?

Here are five follow-up questions:

1. How would you solve for the missing sides if the hypotenuse was unknown?
2. What if angle $A$ was given instead of $B$?
3. How does the Pythagorean theorem relate to these triangle calculations?
4. Can you verify the solutions using the Pythagorean theorem?
5. How do trigonometric ratios change when switching from degrees to radians?

Tip: When solving right triangles, always check if you can use the Pythagorean theorem after finding two sides to ensure accuracy.