Question

solve for side x

triangle:

63 degree
8
x

a. 15.7
b. 17.6
c. 21
d. 15

Answer 1

To solve for side \( x \) in the given triangle, we can use the Law of Sines, which states:

\[
\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}
\]

Given:
- One angle \( A = 63^\circ \)
- One side \( a = 8 \)
- Another side \( b = x \)
- We need to find side \( x \).

To proceed, we first note that we need another angle to find \( x \). However, if we consider the triangle may have two angles, we can apply the logic that if we can find an angle \( B \), we can then find the side \( x \).

### Assume this is a right triangle
In a right triangle, we can use the sine function defined as:

\[
\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}}
\]

However, since we don't have information on other angles or the nature of \( x \) relative to the 63 degrees (for example, whether it's the adjacent or opposite side), let's use the sine law generally.

### Using the Law of Sines
Without the angle opposite to side \( x \), we assume that this is a triangle where the only given angle is \( 63^\circ \) and \( x \) is opposite this angle for calculation purposes.

In a triangle, the remaining angle \( B \) can be calculated with the angle sum property:

\[
A + B + C = 180^\circ
\]

If we assume there's a potential angle \( C \) that makes the triangle valid, and we'll assume for calculations that \( C = 90^\circ \) leading to:

\[
B = 180^\circ - A - C = 180^\circ - 63^\circ - 90^\circ = 27^\circ
\]

Now applying the Law of Sines:
\[
\frac{x}{\sin(63^\circ)} = \frac{8}{\sin(27^\circ)}
\]

Now we can solve for \( x \):
\[
x = \frac{8 \cdot \sin(63^\circ)}{\sin(27^\circ)}
\]

Calculating the sine values:
- \( \sin(63^\circ) \approx 0.8910 \)
- \( \sin(27^\circ) \approx 0.4540 \)

Now we plug in these values:
\[
x = \frac{8 \cdot 0.8910}{0.4540} \approx \frac{7.128}{0.4540} \approx 15.7
\]

Thus, the value of \( x \) is approximately \( 15.7 \). Therefore the answer is:

**a. 15.7**