To find the measures of each side of the triangle $\triangle ABC$, let’s identify and apply the appropriate trigonometric ratios. Given:

• $\angle ABC = 70^\circ$
• $\angle BCA = 90^\circ$
• $AC = 6$ (opposite side to $\angle ABC$)
• We need to find $AB$ (the hypotenuse) and $BC$ (adjacent side to $\angle ABC$).

Let's use trigonometric functions relative to $\angle ABC = 70^\circ$:

1. Finding $BC$ (adjacent side): $\cos(70^\circ) = \frac{BC}{6}$ Solving for $BC$: $BC = 6 \cdot \cos(70^\circ)$

2. Finding $AB$ (hypotenuse): $\sin(70^\circ) = \frac{6}{AB}$ Solving for $AB$: $AB = \frac{6}{\sin(70^\circ)}$

Let me calculate these values.The measures of each side of the triangle are approximately:

• $BC \approx 2.05$
• $AB \approx 6.39$
• $AC = 6$ (as given)

Would you like further details on the solution process, or do you have any questions?

Here are five related questions for further practice:

1. How would the side lengths change if $\angle ABC$ were $60^\circ$ instead of $70^\circ$?
2. Can you calculate the area of this right triangle using the given dimensions?
3. What is the perimeter of $\triangle ABC$ with the side lengths found?
4. How would you find the height from point $C$ to side $AB$?
5. If $AC$ were doubled, how would that affect the other sides?

Tip: In right triangles, knowing one side and an acute angle is enough to determine all other side lengths using trigonometric ratios.