Exercise 10.4 Page: 202

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

**Solution:-**

Steps of construction:

1. Draw a line segment AB = 5.8 cm.

2. At point A, draw a ray P to making an angle of 60^{o} i.e. ∠PAB = 60^{o}.

3. At point B, draw a ray Q to making an angle of 30^{o} i.e. ∠QBA = 30^{o}.

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

Then, 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}.

∴ ∠PQR + ∠QRP + ∠RPQ = 180^{o}

= 105^{o}+ 40^{o}+ ∠RPQ = 180^{o}

= 145^{o} + ∠RPQ = 180^{o}

= ∠RPQ = 180^{o}– 145^{0}

= ∠RPQ = 35^{o}

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

Steps of construction:

1. Draw a line segment PQ = 5 cm.

2. At point P, draw a ray L to making an angle of 105^{o} i.e. ∠LPQ = 35^{o}.

3. At point Q, draw a ray M to making an angle of 40^{o} i.e. ∠MQP = 105^{o}.

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

Then, 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:-**

From the question it is given that,

EF = 7.2 cm

∠E = 110^{o}

∠F = 80^{o}

Now 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^{0}

∠D = -10^{o}

We may observe that the sum of two angles is 190^{o} is greater than 180^{o}. So, it is not possible to construct a triangle.