In this right triangle $\triangle ABC$, we are tasked with solving for $x$, which is the side adjacent to the 70° angle at $C$. Given that:

• $AB = 8$ is the hypotenuse,
• $\angle ACB = 70^\circ$,
• $\angle ABC = 90^\circ$ (right angle).

We can use trigonometric functions to solve for $x$. Since we have the hypotenuse and are looking for the adjacent side to the angle $70^\circ$, we will use the cosine function:

$\cos(70^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{8}$

Solving for $x$:

$x = 8 \cdot \cos(70^\circ)$

Let me calculate that for you.The value of $x$ is approximately $2.7$ when rounded to the nearest tenth.

Would you like more details, or do you have any further questions?

Here are some related questions for further practice:

1. What is the length of the side opposite the $70^\circ$ angle?
2. How would the calculation change if the given angle were $50^\circ$ instead?
3. Can you calculate the area of triangle $ABC$?
4. How would you solve for $x$ using the sine or tangent function?
5. What if the hypotenuse were a different length, say $10$? How would that affect $x$?

Tip: In right triangles, the cosine function is always used to relate the adjacent side to the hypotenuse.