Question

fill in the blank 3 points
$overrightarrow{bd}$ bisects $angle abc$. what is the value of $x$? what is $mangle abd$? what is $mangle abc$?
diagram: point $a$ (left) to $b$ (right), with angle $(10x - 51)^circ$ at $b$; $d$ and $c$ below $b$ with angle $(6x - 11)^circ$ at $b$
$x =$ type your answer...
$mangle abd =$ type your answer... $^circ$
$mangle abc =$ type your answer... $^circ$

Explanation:

Step1: Use Angle Bisector Definition

Since \( \overrightarrow{BD} \) bisects \( \angle ABC \), \( \angle ABD = \angle DBC \). Wait, actually, looking at the diagram, \( \angle ABC \) is split by \( BD \), and the angles given are \( (10x - 51)^\circ \) (for \( \angle ABD \) or maybe the adjacent? Wait, no, the diagram: \( BA \) is a straight line left, \( BD \) is a ray down, \( BC \) is a ray down-right? Wait, no, the angles: \( \angle ABD \) and \( \angle DBC \)? Wait, no, the given angles are \( (10x - 51)^\circ \) for \( \angle ABD \) (between \( BA \) and \( BD \)) and \( (6x - 11)^\circ \) for \( \angle DBC \) (between \( BD \) and \( BC \))? Wait, no, actually, since \( BD \) bisects \( \angle ABC \), so \( \angle ABD = \angle DBC \)? Wait, no, maybe \( \angle ABC \) is the angle between \( BA \) and \( BC \), and \( BD \) bisects it, so \( \angle ABD = \angle DBC \). But the angles given: \( (10x - 51)^\circ \) and \( (6x - 11)^\circ \). Wait, maybe \( \angle ABD = \angle DBC \), so set them equal? Wait, no, maybe the angle \( \angle ABC \) is equal to \( 2 \times \angle ABD \), but looking at the diagram, \( BA \) is a straight line, so maybe \( \angle ABD \) and \( \angle DBC \) are equal? Wait, no, the problem says \( \overrightarrow{BD} \) bisects \( \angle ABC \), so \( \angle ABD = \angle DBC \). Wait, but the angles given are \( (10x - 51)^\circ \) and \( (6x - 11)^\circ \). Wait, maybe I misread. Wait, the angle at \( B \): \( BA \) is horizontal left, \( BD \) is a ray going down, and \( BC \) is a ray going down-right? Wait, no, the angles: \( (10x - 51)^\circ \) is between \( BA \) and \( BD \), and \( (6x - 11)^\circ \) is between \( BD \) and \( BC \). Since \( BD \) bisects \( \angle ABC \), then \( \angle ABD = \angle DBC \)? Wait, no, \( \angle ABC \) is the angle between \( BA \) and \( BC \), so \( \angle ABC = \angle ABD + \angle DBC \), and since \( BD \) bisects it, \( \angle ABD = \angle DBC \). Wait, but the angles given: \( (10x - 51)^\circ \) and \( (6x - 11)^\circ \). Wait, maybe \( \angle ABD = \angle DBC \), so \( 10x - 51 = 6x - 11 \)? Wait, solving that: \( 10x - 6x = 51 - 11 \) → \( 4x = 40 \) → \( x = 10 \). Wait, let's check. If \( x = 10 \), then \( 10x - 51 = 100 - 51 = 49^\circ \), and \( 6x - 11 = 60 - 11 = 49^\circ \). Oh, so they are equal. So that makes sense. So \( x = 10 \). Then \( m\angle ABD = 49^\circ \), and \( m\angle ABC = 2 \times 49^\circ = 98^\circ \).

Step1: Set Angles Equal (Bisector)

Since \( \overrightarrow{BD} \) bisects \( \angle ABC \), \( \angle ABD = \angle DBC \). Thus:
\( 10x - 51 = 6x - 11 \)

Step2: Solve for \( x \)

Subtract \( 6x \) from both sides:
\( 4x - 51 = -11 \)

Add 51 to both sides:
\( 4x = 40 \)

Divide by 4:
\( x = 10 \)

Step3: Find \( m\angle ABD \)

Substitute \( x = 10 \) into \( 10x - 51 \):
\( 10(10) - 51 = 100 - 51 = 49^\circ \)

Step4: Find \( m\angle ABC \)

Since \( BD \) bisects \( \angle ABC \), \( m\angle ABC = 2 \times m\angle ABD \):
\( 2 \times 49 = 98^\circ \)

Answer:

\( x = \boldsymbol{10} \)
\( m\angle ABD = \boldsymbol{49}^\circ \)
\( m\angle ABC = \boldsymbol{98}^\circ \)