Question

From isosceles triangles are shown below. One angle of each is given. Find the other two angles of each.

Answer 1

We know that in an isosceles triangle, two sides are of equal lengths.

Also, angles opposite to equal sides of an isosceles triangle are equal.

(1)

ΔABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ACB = ∠ABC

Using angle sum property in ΔABC, we have:

∠ACB + ∠CBA + ∠BAC = 180°

⇒ ∠BAC + 2∠ACB = 180°

⇒ ∠BAC + 2 × 33° = 180°

⇒ ∠BAC = 180° − 66°

= 114°

Thus, the other two angles of the triangle are of measures 33° and 114°.

(2)

ΔPQR is an isosceles triangle.

∴ PQ = PR

⇒ ∠PRQ = ∠PQR

Using angle sum property in ΔPQR, we have:

∠QRP + ∠RPQ + ∠PQR = 180°

⇒ 2∠PQR + ∠RPQ = 180°

⇒ 2 × 70° + ∠RPQ = 180°

⇒ ∠RPQ = 180° − 140°

= 40°

Thus, the other two angles of the triangle are of measures 70° and 40°.

(3)

ΔSTU is an isosceles triangle.

∴ ST = UT

⇒ ∠TUS = ∠TSU

Using angle sum property in ΔSTU, we have:

∠STU + ∠TUS + ∠TSU = 180°

⇒ 17° + 2∠TSU = 180°

⇒ 2∠TSU = 180° − 17°

⇒∠TSU = = 81.5°

Thus, the other two angles of the triangle are of measures 81.5° each.

(4)

ΔXYZ is an isosceles triangle.

∴ XY = ZY

⇒ ∠YZX = ∠YXZ

Using angle sum property in ΔXYZ, we have:

∠XYZ + ∠YZX + ∠ZXY = 180°

⇒ 115° + 2∠YZX = 180°

⇒ 2∠YZX = 180° − 115°

⇒∠YZX = = 32.5°

Thus, the other two angles of the triangle are of measures 32.5° each.

(5)

ΔABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ACB = ∠ABC

Using angle sum property in ΔABC, we have:

∠ACB + ∠CBA + ∠BAC = 180°

⇒ ∠BAC + 2∠ABC = 180°

⇒ ∠BAC + 2 × 50° = 180°

⇒ ∠BAC = 180° − 100°

= 80°

Thus, the other two angles of the triangle are of measures 80° and 50°.

(6)

ΔPQR is an isosceles triangle.

∴ PQ = PR

⇒ ∠PRQ = ∠PQR

Using angle sum property in ΔPQR, we have:

∠QPR + ∠PRQ + ∠RQP = 180°

⇒ ∠QPR + 2∠RQP = 180°

⇒ ∠QPR + 2 × 67° = 180°

⇒ ∠QPR = 180° − 134°

= 46°

Thus, the other two angles of the triangle are of measures 67° and 46°.