In ΔKLM, start overline, K, M, end overline

Question

KM
is extended through point M to point N, m, angle, L, M, N, equals, left parenthesis, 6, x, plus, 0, right parenthesis, degreesm∠LMN=(6x+0)
∘
, m, angle, M, K, L, equals, left parenthesis, 2, x, plus, 13, right parenthesis, degreesm∠MKL=(2x+13)
∘
, and m, angle, K, L, M, equals, left parenthesis, 2, x, plus, 17, right parenthesis, degreesm∠KLM=(2x+17)
∘
. Find m, angle, K, L, M, .m∠KLM.

Answer 1

To find \( m\angle KLM \) in triangle \( \Delta KLM \) with the angles given, we can use the fact that the sum of the angles in any triangle is \( 180^\circ \). The angles we have are:

1. \( m\angle LMN = 6x \)
2. \( m\angle MKL = 2x + 13 \)
3. \( m\angle KLM = 2x + 17 \)

Since the line segment \( KM \) is extended to point \( N \), we can also establish a relationship between the angles. In triangle \( KLM \), \( m\angle LMN \) is an exterior angle, and it can be expressed as the sum of the two opposite interior angles:

\[ m\angle LMN = m\angle MKL + m\angle KLM \]

Substituting the expressions for the angles, we get:

\[ 6x = (2x + 13) + (2x + 17) \]

Now, let's simplify this equation:

\[ 6x = 2x + 13 + 2x + 17 \]

Combine like terms:

\[ 6x = 4x + 30 \]

Subtract \( 4x \) from both sides:

\[ 2x = 30 \]

Now, divide both sides by 2:

\[ x = 15 \]

Now that we have the value of \( x \), we can find \( m\angle KLM \):

\[ m\angle KLM = 2x + 17 = 2(15) + 17 = 30 + 17 = 47^\circ \]

Thus, the measure of angle \( KLM \) is

\[ \boxed{47^\circ}. \]