Problem-1: In the figure, ∆ABC is a triangle in which ∠ABC = 900, ∠BAC =480 and BD is perpendicular to AC. Find the remaining angles.
Solution: Let remaining angles ∠ABD = x, ∠DBC = y and ∠BCD = z.
∴ ∠ADB = ∠CDB = 900
⟹x=1800 – 1380
Again, ∠ABC = 900
∴ y +x = ∠ABC = 900
⟹ y = 900 – 420
⟹ y = 480
And in ∆BCD,
∠CDB+ y + z = 1800
⟹ 900 + 480 + z = 1800
⟹ z + 1380 = 1800
⟹ z = 1800 – 1380
⟹ z = 420
∴ x=420, y = 480, z = 420.
Problem-2: The vertical angle of an isosceles triangle is 500. Find the other two angles.
Solution: Let, ∆ABC be isosceles triangle where AB = AC and vertical angle ∠BAC = 500. 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 1800.
So, ∠ACB + ∠ABC+ ∠BAC = 1800
⟹ x + x + 500 = 1800
⟹ 2x + 500 = 1800
⟹ 2x = 1800 – 500
⟹ 2x = 1300
⟹ x = 650
∴ ∠ACB = 650 and ∠ABC = 650
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
Again from triangle ADC
Now adding (1) and (2) we have,
⇒(∠BAC+∠DAC)+ ∠ABC+(∠BCA+∠DCA)+ ∠ADC=3600
⇒ ∠BAD+∠A BC+∠BCD+∠ADC=3600
⇒∠A+∠B+∠C+∠D=3600 = 4 right angles. (Proved)