Question

angle addition postulate and angle bisector
7 if (mangle abc = 4x + 16) and (mangle cbx=3x + 6), find the value of (x).
8 if (mangle abc = 5x + 18) and (mangle cbx = 2x+12), find the value of (x), and the (mangle abc).

1. find (mangle klu) if (mangle ulm = 20^{circ}) and (mangle klm = 110^{circ}).
2. find (mangle ija) if (mangle ajk = 61^{circ}) and (mangle ijk = 153^{circ}).

Explanation:

Step1: Apply angle - addition postulate

For problem 7, assume $\angle ABC$ is composed of $\angle CBX$ and another angle. If $\angle ABC = 4x + 16$ and $\angle CBX=3x + 6$, then by angle - addition postulate, we set up the equation $4x+16=3x + 6$.

Step2: Solve the equation for x

Subtract $3x$ from both sides: $4x-3x+16=3x-3x + 6$, which gives $x+16 = 6$. Then subtract 16 from both sides: $x=6 - 16=-10$.

For problem 8:

Step1: Set up the equation

If $\angle ABC = 5x + 18$ and $\angle CBX=2x + 12$, by angle - addition postulate, $5x+18=2x + 12$.

Step2: Solve for x

Subtract $2x$ from both sides: $5x-2x+18=2x-2x + 12$, so $3x+18 = 12$. Subtract 18 from both sides: $3x=12 - 18=-6$. Divide both sides by 3: $x=-2$.

Step3: Find $m\angle ABC$

Substitute $x = - 2$ into $m\angle ABC=5x + 18$, we get $m\angle ABC=5\times(-2)+18=-10 + 18 = 8$.

For problem 9:

Step1: Apply angle - addition postulate

If $\angle KLM=\angle KLU+\angle ULM$, and $\angle ULM = 20^{\circ}$, $\angle KLM = 110^{\circ}$, then $\angle KLU=\angle KLM-\angle ULM$.

Step2: Calculate $\angle KLU$

$\angle KLU=110^{\circ}-20^{\circ}=90^{\circ}$.

For problem 10:

Step1: Apply angle - addition postulate

If $\angle IJK=\angle IJA+\angle AJK$, and $\angle AJK = 61^{\circ}$, $\angle IJK = 153^{\circ}$, then $\angle IJA=\angle IJK-\angle AJK$.

Step2: Calculate $\angle IJA$

$\angle IJA=153^{\circ}-61^{\circ}=92^{\circ}$.

Answer:

1. $x=-10$
2. $x=-2$, $m\angle ABC = 8$
3. $m\angle KLU = 90^{\circ}$
4. $m\angle IJA = 92^{\circ}$