ABC is isosceles with AB = AC, and A = 80. What can we say about B and C? Construct the altitude from A to BC.

In an isosceles triangle, angles opposite to equal sides are also equal. Step 1 — Find angles B and C We are given that triangle ABC is an isosceles triangle. Sides AB and AC are equal. So, the angles opposite to these sides must be equal. This means that angle B is equal to angle C. Let us call this unknown angle . So, B = C = . We know that the sum of all angles in any triangle is 180 degrees. So, we can write an equation for the angles in triangle ABC. We are given that A is 80 degrees. We substitute the known values into our equation. Now, we combine the terms with . To find , we subtract 80 degrees from both sides. To find , we divide 100 degrees by 2. > So, B is 50° and C is 50°. Step 2 — Construct the altitude from A to BC An altitude is a line segment from a vertex. It is drawn perpendicular to the opposite side. Let us draw a line segment from vertex A. This line segment meets the side BC. It forms a 90-degree angle with side BC. Let us name the point where it meets BC as D. So, AD is the altitude from A to BC. In an isosceles triangle, this altitude also bisects the base BC. It also bisects the vertex angle A. Answer > (i) B = 50° and C = 50°. > (ii) The altitude from A to BC is a line segment AD, where D is a point on BC such that AD is perpendicular to BC.

Geometric Twins | IT

Question 29

$\Delta$ ABC is isosceles with AB = AC, and $\angle$ A = 80. What can we say about $\angle$ B and $\angle$ C?

Construct the altitude from A to BC.

Solution

In an isosceles triangle, angles opposite to equal sides are also equal.

Step 1 — Find angles B and C

We are given that triangle ABC is an isosceles triangle. Sides AB and AC are equal. So, the angles opposite to these sides must be equal. This means that angle B is equal to angle C. Let us call this unknown angle $x$ . So, $\angle$ B = $\angle$ C = $x$ . We know that the sum of all angles in any triangle is 180 degrees. So, we can write an equation for the angles in triangle ABC.

$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$

We are given that $\angle$ A is 80 degrees. We substitute the known values into our equation.

$80^\circ + x + x = 180^\circ$

Now, we combine the terms with $x$ .

$80^\circ + 2x = 180^\circ$

To find $2x$ , we subtract 80 degrees from both sides.

$2x = 180^\circ - 80^\circ$

$2x = 100^\circ$

To find $x$ , we divide 100 degrees by 2.

$x = \frac{100^\circ}{2}$

$\boxed{x = 50^\circ}$

So, $\angle$ B is 50° and $\angle$ C is 50°.

Diagram 1

Step 2 — Construct the altitude from A to BC

An altitude is a line segment from a vertex. It is drawn perpendicular to the opposite side. Let us draw a line segment from vertex A. This line segment meets the side BC. It forms a 90-degree angle with side BC. Let us name the point where it meets BC as D. So, AD is the altitude from A to BC. In an isosceles triangle, this altitude also bisects the base BC. It also bisects the vertex angle A.

Answer

(i) $\angle$ B = 50° and $\angle$ C = 50°. (ii) The altitude from A to BC is a line segment AD, where D is a point on BC such that AD is perpendicular to BC.

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