Question

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critical thinking questions:

1. find an angle ( x ) where ( sin x = cos x ).
2. draw and label all three sides of a right triangle that has a ( 40^circ ) angle and a hypotenuse of 10 cm.

Explanation:

Step1: Recall sine definition

For right triangles, $\sin x = \frac{\text{opposite side}}{\text{hypotenuse}}$

Step2: Solve 15)

Identify opp=7, hyp=25. $\sin x = \frac{7}{25}$

Step3: Solve 16)

Identify opp=12, hyp=13. $\sin x = \frac{12}{13}$

Step4: Solve 17)

First find hypotenuse: $\sqrt{6^2+8^2}=10$. Opp=6, hyp=10. $\sin x = \frac{6}{10}=\frac{3}{5}$

Step5: Solve 18)

First find missing leg: $\sqrt{5^2-4^2}=3$. Opp=3, hyp=5. $\sin x = \frac{3}{5}$

Step6: Solve 19)

Identify opp=48, hyp=50. $\sin x = \frac{48}{50}=\frac{24}{25}$

Step7: Solve 20)

Identify opp=60, hyp=61. $\sin x = \frac{60}{61}$

Step8: Solve 21)

Identify opp=24, hyp=25. $\sin x = \frac{24}{25}$

Step9: Solve 22)

Identify opp=16, hyp=34. $\sin x = \frac{16}{34}=\frac{8}{17}$

Step10: Solve 23)

Identify opp=51, hyp=85. $\sin x = \frac{51}{85}=\frac{3}{5}$

Step11: Solve 24)

Identify opp=42, hyp=70. $\sin x = \frac{42}{70}=\frac{3}{5}$

Step12: Solve 25)

Set $\sin x = \cos x$. Use $\sin x = \cos(90^\circ-x)$, so $x=90^\circ-x$. Solve: $2x=90^\circ \implies x=45^\circ$

Step13: Solve 26)

For right triangle with hypotenuse=10 cm, $40^\circ$ angle:

• Opposite side: $10\sin40^\circ\approx6.43$ cm
• Adjacent side: $10\cos40^\circ\approx7.66$ cm

Label hypotenuse=10 cm, opposite≈6.43 cm, adjacent≈7.66 cm to $40^\circ$ angle.

Answer:

1. $\frac{7}{25}$
2. $\frac{12}{13}$
3. $\frac{6}{10}$ or $\frac{3}{5}$
4. $\frac{3}{5}$
5. $\frac{48}{50}$ or $\frac{24}{25}$
6. $\frac{60}{61}$
7. $\frac{24}{25}$
8. $\frac{16}{34}$ or $\frac{8}{17}$
9. $\frac{51}{85}$ or $\frac{3}{5}$
10. $\frac{42}{70}$ or $\frac{3}{5}$
11. $45^\circ$
12. Hypotenuse = 10 cm, Opposite to $40^\circ$: $10\sin40^\circ\approx6.43$ cm, Adjacent to $40^\circ$: $10\cos40^\circ\approx7.66$ cm