Answer :

In first figure,

As seen it seems AB = BC (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠BAC(∠A) = ∠BCA(∠C) = 30°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ ABC,

∠A + ∠B + ∠C = 180°

∴ 30 + ∠B + 30 = 180°

∠B + 60 = 180°

∴ ∠B = 180-60

= 120°

In second figure,

As seen it seems DE = DF (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠DEF(∠E) = ∠DFE(∠F) = y° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ DEF,

∠D + ∠E + ∠F = 180°

∴ 40 + y + y = 180°

∴ 40 + 2y = 180

∴ 2y = 180-40

= 140°

∴

Hence, ∠E = ∠F = 70° (from eq1)

In 3^{rd} figure,

As seen it seems PQ = PR (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠PQR(∠Q) = ∠PRQ(∠R) = y° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ PQR,

∠P + ∠Q + ∠R = 180°

∴ 20 + y + y = 180°

∴ 20 + 2y = 180

∴ 2y = 180-20

= 160°

∴

Hence, ∠Q = ∠R = 80° (from eq1)

In 4^{th} figure,

As seen it seems XY = XZ (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠XYZ(∠Y) = ∠XZY(∠Z) = m° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ XYZ,

∠X + ∠Y + ∠Z = 180°

∴ 100 + m + m = 180°

∴ 100 + 2m = 180

∴ 2m = 180-100

= 80

∴

Hence, ∠Y = ∠Z = 40° (from eq1)

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<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I