**1. Construct ΔABC, given m ∠A =60 ^{O}, m ∠B = 30^{O} and AB = 5.8 cm.**

**Solution:**

Steps:

Step 1. Draw a line segment AB = 5.8 cm.

Step 2. Draw a ray P, at point A to make an angle of 60^{O} i.e. ∠PAB = 60^{O}.

Step 3. Draw a ray Q, at point B to make an angle of 30^{O} i.e. ∠QBA = 30^{O}.

Step 4. Now the two rays AP and BQ intersect each other at point C.

Hence, ΔABC is the required triangle.

**2. Construct ΔPQR if PQ = 5 cm, m∠PQR = 105 ^{o} and m∠QRP = 40^{o}**.

(Hint: Recall angle-sum property of a triangle).

**Solution:**

We know that the sum of the angles of a triangle is 180^{o}.

Now,

∠PQR + ∠QRP + ∠RPQ = 180^{o}

105^{o}+ 40^{o}+ ∠RPQ = 180^{o}

145^{o} + ∠RPQ = 180^{o}

∠RPQ = 180^{o }– 145^{o}

∠RPQ = 35^{o}

Hence, the measures of ∠RPQ is 35^{o}.

Steps:

Step 1. Draw a line segment PQ = 5 cm.

Step 2. Draw a ray L, at point P to make an angle of 105^{o} i.e. ∠LPQ = 35^{o}.

Step 3. Draw a ray M, at point Q to make an angle of 40^{o} i.e. ∠MQP = 105^{o}.

Step 4. Now the two rays PL and QM intersect each other at the point R.

Hence, ΔPQR is the required triangle.

**3. Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E = 110°, and** **m∠F = 80°. Justify your answer.**

**Solution:**

Given,

EF = 7.2 cm

∠E = 110^{o}

∠F = 80^{o}

Now,

According to the question we have to check whether it is possible to construct ΔDEF from the given values.

We know that the sum of the angles of a triangle is 180^{o}.

Then,

∠D + ∠E + ∠F = 180^{o}

∠D + 110^{o}+ 80^{o}= 180^{o}

∠D + 190^{o} = 180^{o}

∠D = 180^{o }– 190^{o}

∠D = -10^{o}

Now, we observe that the sum of the two angles is 190^{o}, which is greater than 180^{o}.

Hence, it is not possible to construct a triangle.

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