Question: Question 4 — give a reason for each step.
A triangle has an apex angle of $50^\circ$ and two equal base angles.
Find each base angle, giving your reason.
The two base angles are $65^\circ$. Because it is an isosceles triangle, meaning two equal sides, both base angles must be equal. All angles in a $\triangle$ = $180^\circ$, so $180-50 = 130$, $130 \div 2 = 65$.
(b) In triangle $ABC$, a line $DE$ is drawn parallel to the base $BC$ ($D$ on $AB$, $E$ on $AC$). Angle $ABC = 65^\circ$.
Find angle $BDE$ (marked ?), giving a reason for each step.
Angle $BDE$ would equal $115^\circ$. This is because angle $BDE$ & $DBC$ are both marked between parallel lines, making them co-interior angles. These angle = $180^\circ$ as they are on a $\text{─}$ so $180-65 = 115$.
(c) In triangle $ABC$, angle $A = 55^\circ$ and angle $B = 65^\circ$. $CD$ is drawn perpendicular to $AB$. Find angle $ACD$ and angle $BCD$, justifying each step, then show they agree with the whole angle $ACB$.
Angle $ACD$ equals $35^\circ$ because I know a right angle equals $90^\circ$ and all angles in a triangle equal $180^\circ$. Therefore $55 + 90 = 145$, & $180-145$ equals $35$. Angle
$BCD$ equals $25^\circ$ for the same reason. From this I can gather that angle $ACB$ equals $60^\circ$, because it is angle $ACD + BCD$. In conclusion, this agrees with $ACB$ by being $65+55+60 = 180$.
1. **Problem Statement:**
- (a) Find each base angle of an isosceles triangle with apex angle $50^\circ$.
- (b) In $\triangle ABC$, with $DE \parallel BC$, find angle $BDE$ given $\angle ABC = 65^\circ$.
- (c) In $\triangle ABC$, with $\angle A = 55^\circ$, $\angle B = 65^\circ$, and $CD \perp AB$, find angles $ACD$ and $BCD$ and verify their sum equals $\angle ACB$.
2. **Step (a): Find base angles of isosceles triangle**
- Formula: Sum of angles in a triangle is $180^\circ$.
- Since apex angle is $50^\circ$, sum of base angles is:
$$180 - 50 = 130$$
- Because the triangle is isosceles, base angles are equal:
$$\text{Each base angle} = \frac{130}{2} = 65$$
- **Reason:** Isosceles triangle has two equal sides, so base angles opposite those sides are equal.
3. **Step (b): Find angle $BDE$ with $DE \parallel BC$**
- Given $\angle ABC = 65^\circ$.
- $DE$ is parallel to $BC$, so angles $BDE$ and $DBC$ are co-interior angles.
- Co-interior angles sum to $180^\circ$:
$$\angle BDE + \angle DBC = 180$$
- Substitute $\angle DBC = 65^\circ$:
$$\angle BDE = 180 - 65 = 115$$
- **Reason:** Co-interior angles between parallel lines sum to $180^\circ$.
4. **Step (c): Find angles $ACD$ and $BCD$ with $CD \perp AB$**
- $CD$ is perpendicular to $AB$, so $\angle CDB = 90^\circ$.
- In right triangle $ACD$:
$$\angle A + \angle ACD + 90 = 180$$
$$55 + \angle ACD + 90 = 180$$
$$\angle ACD = 180 - 145 = 35$$
- In right triangle $BCD$:
$$\angle B + \angle BCD + 90 = 180$$
$$65 + \angle BCD + 90 = 180$$
$$\angle BCD = 180 - 155 = 25$$
- Sum angles $ACD$ and $BCD$:
$$35 + 25 = 60$$
- Check sum of angles in $\triangle ABC$:
$$55 + 65 + 60 = 180$$
- **Reason:** Angles in a triangle sum to $180^\circ$, and perpendicular lines create right angles.
**Final answers:**
- Base angles: $65^\circ$
- Angle $BDE$: $115^\circ$
- Angles $ACD = 35^\circ$, $BCD = 25^\circ$, sum $60^\circ$ equals $\angle ACB$.