Question: User: Top-left: right triangle with a vertical leg labeled 15 cm, a horizontal base labeled a = ?, hypotenuse labeled c = ?, and angle $\theta = 45^\circ$ at the bottom-right corner.
Top-right: Calculate the length of $c$.
Top-right: Calculate the length of $a$.
Bottom-left: #3. right triangle with base labeled $a = ?$, vertical leg labeled $b = ?$, hypotenuse labeled $c = 8$ inches, and angle $\theta = 40^\circ$ at the bottom-left corner.
Bottom-right: Calculate the length of $a$.
Bottom-right: Calculate the length of $b$.
1. **Problem 1:** Right triangle with vertical leg $15$ cm, angle $\theta = 45^\circ$ at bottom-right corner, find hypotenuse $c$ and base $a$.
2. **Step 1:** Identify sides relative to angle $\theta=45^\circ$.
- Opposite side to $\theta$ is vertical leg $15$ cm.
- Adjacent side is base $a$.
- Hypotenuse is $c$.
3. **Step 2:** Use trigonometric ratios for $45^\circ$.
- $\sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{c}$
- $\cos 45^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{c}$
4. **Step 3:** Calculate $c$ using $\sin 45^\circ = \frac{15}{c}$.
- $\sin 45^\circ = \frac{\sqrt{2}}{2}$
- So, $\frac{\sqrt{2}}{2} = \frac{15}{c}$
- Multiply both sides by $c$: $c \cdot \frac{\sqrt{2}}{2} = 15$
- Divide both sides by $\frac{\sqrt{2}}{2}$:
$$c = \frac{15}{\frac{\sqrt{2}}{2}} = 15 \times \frac{2}{\sqrt{2}}$$
- Simplify:
$$c = 15 \times \frac{2}{\sqrt{2}} = 15 \times \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 15 \times \frac{2\sqrt{2}}{2} = 15 \sqrt{2}$$
5. **Step 4:** Calculate $a$ using $\cos 45^\circ = \frac{a}{c}$.
- $\cos 45^\circ = \frac{\sqrt{2}}{2}$
- So, $\frac{\sqrt{2}}{2} = \frac{a}{15\sqrt{2}}$
- Multiply both sides by $15\sqrt{2}$:
$$a = 15\sqrt{2} \times \frac{\sqrt{2}}{2}$$
- Simplify:
$$a = 15 \times \frac{\sqrt{2} \times \sqrt{2}}{2} = 15 \times \frac{2}{2} = 15$$
6. **Answer for Problem 1:**
- Hypotenuse $c = 15\sqrt{2}$ cm
- Base $a = 15$ cm
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7. **Problem 2:** Right triangle with hypotenuse $c=8$ inches, angle $\theta=40^\circ$ at bottom-left corner, find base $a$ and vertical leg $b$.
8. **Step 1:** Identify sides relative to angle $\theta=40^\circ$.
- Adjacent side to $\theta$ is base $a$.
- Opposite side is vertical leg $b$.
- Hypotenuse is $c=8$ inches.
9. **Step 2:** Use trigonometric ratios.
- $\cos 40^\circ = \frac{a}{8}$
- $\sin 40^\circ = \frac{b}{8}$
10. **Step 3:** Calculate $a$.
- $a = 8 \cos 40^\circ$
- $\cos 40^\circ \approx 0.7660$
- So, $a \approx 8 \times 0.7660 = 6.128$
11. **Step 4:** Calculate $b$.
- $b = 8 \sin 40^\circ$
- $\sin 40^\circ \approx 0.6428$
- So, $b \approx 8 \times 0.6428 = 5.142$
12. **Answer for Problem 2:**
- Base $a \approx 6.13$ inches
- Vertical leg $b \approx 5.14$ inches