Let's go ahead and solve these triangle problems step by step, using trigonometric principles for each.

Problem 3

For the right triangle:

• Angle = 67°
• Opposite side = 29
• Hypotenuse = x

We can use the sine function, which relates the opposite side to the hypotenuse:

$\sin(\text{Angle}) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

$\sin(67^\circ) = \frac{29}{x}$

Solve for $x$:

$x = \frac{29}{\sin(67^\circ)}$

Problem 4

For this right triangle:

• Angle = 29°
• Opposite side = x
• Adjacent side = 12

We will use the tangent function, which relates the opposite side to the adjacent side:

$\tan(\text{Angle}) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(29^\circ) = \frac{x}{12}$

Solve for $x$:

$x = 12 \times \tan(29^\circ)$

Problem 7

For this triangle:

• Angle = 51°
• Opposite side = x
• Hypotenuse = 15

Again, we use the sine function:

$\sin(51^\circ) = \frac{x}{15}$

Solve for $x$:

$x = 15 \times \sin(51^\circ)$

Problem 8

For this triangle:

• Angle = 37°
• Opposite side = x
• Hypotenuse = AB (unknown)

Using sine again:

$\sin(37^\circ) = \frac{x}{AB}$

This gives the relationship, but we would need the length of AB to calculate $x$.

Would you like the exact numerical answers for these, or is there a particular step you'd like more detail on? Let me know!