Question

Find the value of x
A=21degrees
B=16x+49degrees
C=55degrees
X=

Answer 1

Find the value of x
A=3x+30degreea
B=1xdegrees

Answer 2

Is 90 degree

Answer 3

To find the value of \( x \) in triangle \( ABC \), we can use the property that the sum of the angles in a triangle equals \( 180 \) degrees. Therefore, we have the equation:

\[
A + B + C = 180
\]

Substituting the values for \( A \), \( B \), and \( C \):

\[
21 + (16x + 49) + 55 = 180
\]

Now, simplify the equation:

\[
21 + 49 + 55 + 16x = 180
\]

Calculating the constant terms:

\[
125 + 16x = 180
\]

Next, isolate \( 16x \):

\[
16x = 180 - 125
\]

\[
16x = 55
\]

Now, divide by 16 to solve for \( x \):

\[
x = \frac{55}{16}
\]

Simplifying the fraction yields:

\[
x = 3.4375
\]

Thus, the value of \( x \) is:

\[
\boxed{3.4375}
\]

Answer 4

To find the value of \( x \) in the triangle where \( A = 3x + 30^\circ \) and \( B = 1x \) degrees, we need to remember that the sum of the angles in a triangle is equal to \( 180^\circ \).

Assuming \( C \) as the third angle in the triangle, we can write the equation as:

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

Substituting for \( A \) and \( B \):

\[
(3x + 30) + x + C = 180
\]

To isolate \( C \), we rearrange the equation:

\[
4x + 30 + C = 180
\]

Now, let's solve for \( C \):

\[
C = 180 - (4x + 30)
\]

Thus,

\[
C = 180 - 4x - 30
\]
\[
C = 150 - 4x
\]

Now, since angles cannot be negative, we have:

1. \( A = 3x + 30 > 0 \)
2. \( B = 1x > 0 \)
3. \( C = 150 - 4x > 0 \)

Let’s set up the inequalities.

### For Angle A:
\[
3x + 30 > 0 \implies 3x > -30 \implies x > -10
\]

### For Angle B:
\[
x > 0
\]

### For Angle C:
\[
150 - 4x > 0 \implies 150 > 4x \implies x < 37.5
\]

### Summary of inequalities:
- \( x > 0 \)
- \( x < 37.5 \)

Now, if we assume there is no third angle, we can use

\[
3x + 30 + x = 180
\]

Combining like terms gives us the new equation:

\[
4x + 30 = 180
\]

Now let's solve for \( x \):

\[
4x = 180 - 30
\]
\[
4x = 150
\]
\[
x = \frac{150}{4} = 37.5
\]

However, \( x \) cannot be \( 37.5 \) as that does not satisfy \( x < 37.5 \).

We conclude that for the angle conditions \( A \), \( B \), and \( C \), they sum up to \( 180^\circ \) and must simultaneously satisfy the above inequalities.

Thus, the only possible value, while respecting \( x \) remains less than \( 37.5 \) would be:

\[ x = 30 \]

Please verify that \(x\) fits the angle constraints. The general formula \(x < 37.5\) allows values in the range \(0 < x < 37.5\). Therefore \(x = 30\) is valid within that range.

Thus, the value of \( x \) is:

\[
\boxed{30}
\]

Answer 5

If you are asking whether any of the angles can be equal to \(90^\circ\) in the context of the triangle you proposed, it depends on the value of \(x\).

Let's evaluate when the angles \(A\) and \(B\) could or could not equal \(90^\circ\) given:

- \( A = 3x + 30^\circ \)
- \( B = 1x \)

### For Angle A:
To find out when \( A = 90^\circ \):

\[
3x + 30 = 90
\]
\[
3x = 90 - 30
\]
\[
3x = 60
\]
\[
x = 20
\]

### For Angle B:
To find out when \( B = 90^\circ \):

\[
1x = 90
\]
\[
x = 90
\]

However, \(x = 90\) does not satisfy the earlier established inequality \( x < 37.5 \) from the triangle's angle constraints.

### Summary:
- \( A = 90^\circ \) when \( x = 20 \)
- \( B = 90^\circ \) when \( x = 90 \), but this is not valid in the context of a triangle since it exceeds the previous limits.

Thus:
- If you choose \( x = 20 \), then angle \( A \) would be \( 90^\circ \), and consequently, angle \( B \) would be:
\[
B = 20^\circ
\]
yielding:
\[
C = 180 - 90 - 20 = 70^\circ
\]

In conclusion, only angle \( A \) can be \( 90^\circ \) in this configuration, and \( x = 20 \) satisfies the triangle's angle constraints. If that's what you were pointing to, please confirm or specify further!