## Problem-1: In the figure, ∆ABC is a triangle in which ∠ABC = 90^{0}, ∠BAC =48^{0} and BD is perpendicular to AC. Find the remaining angles.

**Solution**: Let remaining angles ∠ABD = x, ∠DBC = y and ∠BCD = z.

Since BD⊥AC

∴ ∠ADB = ∠CDB = 90^{0}

x+ ∠ADB+48^{0}=180^{0}

⟹x+90^{0}+48^{0}=180^{0}

⟹x+138^{0}=180^{0}

⟹x=180^{0} – 138^{0}

⟹x=42^{0}.

Again, ∠ABC = 90^{0}

∴ y +x = ∠ABC = 90^{0}

⟹ y+42^{0}=90^{0}

⟹ y = 90^{0} – 42^{0}

⟹ y = 48^{0}

And in ∆BCD,

∠CDB+ y + z = 180^{0}

⟹ 90^{0} + 48^{0} + z = 180^{0}

⟹ z + 138^{0} = 180^{0}

⟹ z = 180^{0} – 138^{0}

⟹ z = 42^{0}

∴ x=42^{0}, y = 48^{0}, z = 42^{0}.

### Problem-2: The vertical angle of an isosceles triangle is 50^{0}. Find the other two angles.

**Solution**: Let, ∆ABC be isosceles triangle where AB = AC and vertical angle ∠BAC = 50^{0}. We have to find out the values of other two angles ∠ABC and ∠ACB.

Since in the ∆ABC, AB = AC

∴∠ACB = ∠ABC [opposite angles are of equal arms are equal]

Let, ∠ACB = ∠ABC = x.

Now, we know that the sum of the three angles of a triangle is 180^{0}.

So, ∠ACB + ∠ABC+ ∠BAC = 180^{0}

⟹ x + x + 50^{0} = 180^{0}

⟹ 2x + 50^{0} = 180^{0}

⟹ 2x = 180^{0} – 50^{0}

⟹ 2x = 130^{0}

⟹ x = 65^{0}

∴ ∠ACB = 65^{0} and ∠ABC = 65^{0}

# Problem-3: Prove that the sum of the angles of a quadrilateral is equal to 4 right angles.

**Solution**: General enunciation: We have to prove that the sum of the angles of a quadrilateral is equal to 4 right angles.

Particular enunciation: Let four angles of a quadrilateral ABCD be ∠A, ∠B, ∠C, and ∠D. We have to prove that sum of the four angles equal to 4 right angles that is ∠A+∠B+∠C+∠D= 4 right angles.

Construction: We join A with C and B with D.

Proof: We know the sum of the three angles of a triangle is equal to two right angles.

From triangle ABC

∠BAC+∠ABC+∠BCA=180^{0}—————————————— (1)

Again from triangle ADC

∠DAC+∠ADC+∠DCA=180^{0}—————————————— (2)

Now adding (1) and (2) we have,

∠BAC+∠ABC+∠BCA+∠DAC+∠ADC+∠DCA=360^{0}

⇒(∠BAC+∠DAC)+ ∠ABC+(∠BCA+∠DCA)+ ∠ADC=360^{0}

⇒ ∠BAD+∠A BC+∠BCD+∠ADC=360^{0}

⇒∠A+∠B+∠C+∠D=360^{0 }= 4 right angles. (Proved)