The document RD Sharma Solutions - Ex-15.3, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**1. âˆ CBX is an exterior angle of Î”ABC at B. Name**

**(i) the interior adjacent angle**

**(ii) the interior opposite angles to exterior âˆ CBX**

**Also, name the interior opposite angles to an exterior angle at A.**

**Solution.**

(i) âˆ ABC

(ii) âˆ BAC and âˆ ACB

Also the interior angles opposite to exterior are âˆ ABC and âˆ ACB

**2. In the fig, two of the angles are indicated. What are the measures of âˆ ACX and âˆ ACB?**

**Solution.**

In Î”ABC, âˆ A = 50Â° and âˆ B = 55Â°

Because of the angle sum property of the triangle, we can say that

âˆ A+âˆ B+âˆ C = 180Â°

50Â° +55Â° +âˆ C = 180Â°

Or

âˆ C=75Â°

âˆ ACB = 75Â°

âˆ ACX = 180Â°âˆ’âˆ ACB = 180Â°âˆ’75Â° = 105Â°

**3. In a triangle, an exterior angle at a vertex is 95Â° and its one of the interior opposite angles is 55Â°. Find all the angles of the triangle.**

**Solution.**

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

Hence, for the given triangle, we can say that :

âˆ ABC+ âˆ BAC = âˆ BCO

55Â° + âˆ BAC = 95Â°

Or,

âˆ BAC = 95Â°â€“ 95Â°

= âˆ BAC = 40Â°

We also know that the sum of all angles of a triangle is 180Â°.

Hence, for the given Î”ABC, we can say that :

âˆ ABC + âˆ BAC + âˆ BCA = 180Â°

55Â° + 40Â° + âˆ BCA = 180Â°

Or,

âˆ BCA =180Â° â€“95Â°

= âˆ BCA = 85Â°

**4. One of the exterior angles of a triangle is 80Â°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?**

**Solution.**

Let us assume that A and B are the two interior opposite angles.

We know that âˆ A is equal to âˆ B.

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

Hence, we can say that :

âˆ A + âˆ B = 80Â°

Or,

âˆ A +âˆ A = 80Â° (âˆ A= âˆ B)

2âˆ A = 80Â°

âˆ A = 40Â°2 =40Â°

âˆ A= âˆ B = 40Â°

Thus, each of the required angles is of 40Â°.

**5. The exterior angles, obtained on producing the base of a triangle both ways are 104Â° and 136Â°. Find all the angles of the triangle.**

**Solution.**

In the given figure, âˆ ABE and âˆ ABC form a linear pair.

âˆ ABE + âˆ ABC =180Â°

âˆ ABC = 180Â°â€“ 136Â°

âˆ ABC = 44Â°

We can also see that âˆ ACD and âˆ ACB form a linear pair.

âˆ ACD + âˆ ACB = 180Â°

âˆ AUB= 180Â°â€“ 104Â°

âˆ ACB = 76Â°

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

Therefore, we can say that :

âˆ BAC + âˆ ABC =104Â°

âˆ BAC = 104Â°â€“ 44Â° = 60Â°

Thus,

âˆ ACE = 76Â°

and

âˆ BAC = 60Â°

**6. In Fig, the sides BC, CA and BA of a Î”ABC have been produced to D, E and F respectively. If âˆ ACD=105Â° and âˆ EAF=45Â° ; find all the angles of the Î”ABC**

**Solution.**

In a Î”ABC, âˆ BAC and âˆ EAF are vertically opposite angles.

Hence, we can say that :

âˆ BAC= âˆ EAF = 45Â°

Considering the exterior angle property, we can say that :

âˆ BAC + âˆ ABC = âˆ ACD = 105Â°

âˆ ABC = 105Â°â€“ 45Â° = 60Â°

Because of the angle sum property of the triangle, we can say that :

âˆ ABC+ âˆ ACS +âˆ BAC = 180Â°

âˆ ACB = 75Â°

Therefore, the angles are 45Â°, 65Â° and 75Â°.

**7. In Fig, AC perpendicular to CE and C âˆ A:âˆ B:âˆ C=3:2:1. Find the value of âˆ ECD.**

**Solution.**

In the given triangle, the angles are in the ratio 3 : 2 : 1.

Let the angles of the triangle be 3x, 2x and x.

Because of the angle sum property of the triangle, we can say that :

3x+2x+x = 180Â°

6x = 180Â°

Or,

x = 30Â° â€¦ (i)

Also, âˆ ACB + âˆ ACE + âˆ ECD = 180Â°

x+ 90Â° + âˆ ECD = 180Â° (âˆ ACE = 90Â° )

âˆ ECD= 60Â° [From (i)]

**8 A student when asked to measure two exterior angles of Î”ABC observed that the exterior angles at A and B are of 103Â° and 74Â° respectively. Is this possible? Why or why not?**

Here,

**Solution.**

Internal angle at A+ External angle at A = 180Â°

Internal angle at A + 103Â° =180Â°

Internal angle at A = 77Â°

Internal angle at B + External angle at B = 180Â°

Internal angle at B + 74Â° = 180Â°

Internal angle at B = 106Â°

Sum of internal angles at A and B = 77Â° + 106Â° =183Â°

It means that the sum of internal angles at A and B is greater than 180Â°, which cannot be possible.

**9. In Fig, AD and CF are respectively perpendiculars to sides BC and AB of Î”ABC. If âˆ FCD=50Â°, find âˆ BAD**

**Solution.**

We know that the sum of all angles of a triangle is 180Â°

Therefore, for the given Î”FCB, we can say that :

âˆ FCB+ âˆ CBF+ âˆ BFC= 180Â°

50Â° + âˆ CBF + 90Â°= 180Â°

Or,

âˆ CBF = 180Â° â€“50Â°â€“ 90Â° = 40Â° â€¦ (i)

Using the above rule for Î”ABD, we can say that :

âˆ ABD + âˆ BDA + âˆ BAD = 180Â°

âˆ BAD = 180Â° â€“90Â°â€“ 40Â° = 50Â° [from (i)]

**10. In Fig, measures of some angles are indicated. Find the value of x.**

**Solution.**

Here,

âˆ AED + 120Â° = 180Â° (Linear pair)

âˆ AED = 180Â°â€“ 120Â° = 60Â°

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ADE, we can say that :

âˆ ADE + âˆ AED + âˆ DAE = 180Â°

60Â°+ âˆ ADE + 30Â° =180Â°

Or,

âˆ ADE =180Â°â€“ 60Â°â€“ 30Â° = 90Â°

From the given figure, we can also say that :

âˆ FDC + 90Â° = 180Â° (Linear pair)

âˆ FDC = 180Â°â€“ 90Â° = 90Â°

Using the above rule for Î”CDF, we can say that :

âˆ CDF + âˆ DCF + âˆ DFC = 180Â°

90Â° + âˆ DCF + 60Â° =180Â°

âˆ DCF = 180Â°âˆ’60Â°âˆ’90Â°= 30Â°

Also,

âˆ DCF + x = 180Â° (Linear pair)

30Â° + x = 180Â°

Or,

x = 180Â°â€“ 30Â° = 150Â°