Question

9.
a c b
(3x + 50)° (x + 10)°
d

____ = x
____ = m∠acd
____ = m∠bcd
12

Explanation:

Step1: Use angle - sum property

Since $\angle ACB$ is a straight - line angle and $\angle ACB=\angle ACD+\angle BCD$, and $\angle ACB = 180^{\circ}$, we have the equation $(3x + 50)+(x + 10)=180$.

Step2: Combine like terms

Combine the $x$ terms and the constant terms on the left - hand side of the equation: $3x+x+50 + 10=180$, which simplifies to $4x+60 = 180$.

Step3: Isolate the variable term

Subtract 60 from both sides of the equation: $4x+60-60=180 - 60$, resulting in $4x=120$.

Step4: Solve for $x$

Divide both sides of the equation by 4: $\frac{4x}{4}=\frac{120}{4}$, so $x = 30$.

Step5: Find $\angle ACD$

Substitute $x = 30$ into the expression for $\angle ACD$: $m\angle ACD=3x + 50=3\times30+50=90 + 50=140^{\circ}$.

Step6: Find $\angle BCD$

Substitute $x = 30$ into the expression for $\angle BCD$: $m\angle BCD=x + 10=30+10=40^{\circ}$.

Answer:

$x = 30$, $m\angle ACD=140^{\circ}$, $m\angle BCD=40^{\circ}$