Chapter 6: Triangles

Solution:-

(c) 120^o

We know that, exterior angle is equal to sum of opposite interior angles.

From the figure,

∠ACD = ∠A + ∠B

∠ACD = 25^o + 35^o

= 60°

Then, in another triangle

x is exterior angle

∴ x = 60° + ∠ACD

x = 60° + 60°

x = 120°

Solution: –

(c) 360^o

From the figure, we can find that there are two triangles.

So, consider the ΔABC,

We know that, sum of the interior angles of the triangle is equal to 180^o.

Therefore, ∠A + ∠B + ∠C = 180^o

Now, consider the ΔDEF,

∠D + ∠E + ∠F = 180^o

Then,

= ∠A + ∠B + ∠C + ∠D + ∠E + ∠F

= 180^o + 180^o

= 360^o

Solution:-

(b) 40^o

Consider the ΔPQR.

From the exterior angle property

∠RPU = ∠PRQ + ∠PQR

140^O = 2 ∠PQR … [given PQ = PR]

∠PQR = 140/2

∠PQR = 70^o

Given, ST || QR and QS is transversal.

From the property of corresponding angles, ∠PST = ∠PQR = 70^o

Now, consider the ΔQSR

RS = RQ … [from the question]

So, ∠SQR = ∠RSQ = 70^O

Then, PQ is a straight line.

∠PST + ∠TSR + ∠RSQ = 180^o

70^o + ∠TSR + 70^o = 180^o

140^o + ∠TSR = 180^o

∠TSR = 180^o – 140^o

∠TSR = 40^o

Solution:-

(a) 50^o

From the question it is given that, ∠BAC = 90^o AD ⊥ BC and ∠BAD = 50^o

So, ∠DAC = ∠BAC – ∠BAD

= 90^o – 50^o

= 40^o

The, consider the ΔADC

From the rule of exterior angle property = ∠ADB = ∠DAC + ∠ACD

90^o = 40^o + ∠ACD

∠ACD = 90 – 40

∠ACD = 50^o

Solution:-

(c) right

Solution:-

(b) 65^o

Let us assume the interior opposite angles are Q and Q.

Then, 130^o = Q + Q … [from exterior angle property]

2Q = 130^o

Q = 130^o/2

Q = 65^o

Therefore, the measure of each interior opposite angle is 65^o.

Solution: –

(d) 145^o

From the question it is given that, one of the angles of triangle is 110^o

We know that, sum of all angles of triangle is equal to 180^o.

So, sum of other 2 angles is 180^o – 110^o = 70^o

Then, both angles get halved 70^o/2 = 35^o

Sum of bisected angles will be half of sum of angles of triangle.

Then, the third angle will be = 180^o – 35^o

= 145^o

Solution: –

(b) BE

We know that the median of triangle divides the opposite side into two equal parts.

Hence, BE is the median (as AE = EC).

Solution:-

(c) 100^o

As we know that, exterior angle is sum of opposite interior angles.

Then, the exterior angle formed by producing QR

It has opposite interior angles ∠P and ∠Q

Therefore, exterior angle = ∠P + ∠Q

= 60^o + 40^o

= 100^o

Solution: –

(d) 40°, 132°, 18°

We know that, sum of angles of triangle is equal to 180^o.

But, 40^o + 132^o + 18^o = 190

So, these triplets cannot be the angles of a triangle.