Answer :

On the basis of the question, the figure can be drawn as follows:

We know that in a triangle sum of all its angles is 180^{o}

∴ In triangle ABC, we have

∠ A + ∠ B + ∠ C = 180^{o}

105^{o} + 45^{o} + ∠ C = 180^{o}

∠ C = 180^{o} – 150^{o}

∠ C = 30^{o}

**Steps of construction:**

(i) Firstly, draw a triangle ABC having side BC = 7 cm, ∠ B = 45^{o} and ∠ C = 30^{o}

(ii) Now draw a BX by making an acute angle with the side BC on the opposite side of the vertex A

(iii) After this locate 4 points on BX i.e. B_{1}, B_{2}, B_{3} and B_{4}

(iv) Now, join B_{3}C and then draw a parallel line through B_{4} to B_{3}C which intersect the extended BC at C’

(v) From C’ draw a line parallel to AC which intersects the extended line segment at C’

∴

**Justification:**

From the figure, we have:

As line A’C’ is parallel to Ac, therefore, it will make the same angle with the line BC

∴ ∠ A’C’B = ∠ ACB (Corresponding angles are equal) (i)

Now, in triangle A’B’C’ and triangle ABC we have:

∠ B = ∠ B (Common)

∠ A’C’B = ∠ ACB [From equation (i)]

∴ (By AA similarity rule)

As the corresponding sides of both the similar triangles are in the same ratio

∴

Hence, the construction is justified.

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