## RD Sharma Solutions Class 7 Chapter 15 Exercise 15.3

#### Exercise 15.3

*1. \(\angle CBX\) is an exterior angle of \(\triangle ABC\) at B. Name *

*(i) the interior adjacent angle *

*(ii) the interior opposite angles to exterior \(\angle CBX\) *

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

(i) \(\angle ABC\)

(ii) \(\angle BAC and \angle ACB\)

Also the interior angles opposite to exterior are \(\angle ABC\)

*2. In the fig, two of the angles are indicated. What are the measures of \(\angle ACX\) and \(\angle ACB\)?*

In \(\triangle ABC\)

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

\(\angle A + \angle B+\angle C=180Â°\)

\(50Â°\)

Or

\(\angle C=75Â°\)

\(\angle ACB=75Â°\)

\(\angle ACX=180Â°-\angle ACB=180Â°-75Â°=105Â°\)

*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. *

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

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

\(\angle ABC\)

\(55Â°\)

Or,

\(\angle BAC\)

= \(\angle BAC\)

We also know that the sum of all angles of a triangle is \(180Â°\)

Hence, for the given \(\triangle ABC\)

\(\angle ABC\)

\(55Â°\)

Or,

\(\angle BCA\)

= \(\angle BCA\)

*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? *

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

We know that \(\angle A\)

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

Hence, we can say that :

\(\angle A\)

Or,

\(\angle A\)

2\(\angle A\)

\(\angle A\)

\(\angle A\)

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. *

In the given figure, \(\angle ABE\)

\(\angle ABE\)

\(\angle ABC\)

\(\angle ABC\)

We can also see that \(\angle ACD\)

\(\angle ACD\)

\(\angle AUB\)

\(\angle ACB\)

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

Therefore, we can say that :

\(\angle BAC\)

\(\angle BAC\)

Thus,

\(\angle ACE\)

and

\(\angle BAC\)

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

In a \(\triangle ABC\)

Hence, we can say that :

\(\angle BAC\)

Considering the exterior angle property, we can say that :

\(\angle BAC\)

\(\angle ABC\)

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

\(\angle ABC\)

\(\angle ACB\)

Therefore, the angles are \(45Â°\)

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

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 = \(30Â°\)

Also, \(\angle ACB\)

x+ \(90Â°\)

\(\angle ECD\)

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

Here,

Internal angle at A+ External angle at A = \(180Â°\)

Internal angle at A + \(103Â°\)

Internal angle at A = \(77Â°\)

Internal angle at B + External angle at B = \(180Â°\)

Internal angle at B + \(74Â°\)

Internal angle at B = \(106Â°\)

Sum of internal angles at A and B = \(77Â°\)

It means that the sum of internal angles at A and B is greater than \(180Â°\)

*9. In Fig, AD and CF are respectively perpendiculars to sides BC and AB of \(\triangle ABC\). If \(\angle FCD=50Â°\), find \(\angle BAD\)*

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

Therefore, for the given \(\triangle FCB\)

\(\angle FCB\)

\(50Â°\)

Or,

\(\angle CBF\)

Using the above rule for \(\triangle ABD\)

\(\angle ABD\)

\(\angle BAD\)

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

Here,

\(\angle AED\)

\(\angle AED\)

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

Therefore, for \(\triangle ADE\)

\(\angle ADE\)

\(60Â°\)

Or,

\(\angle ADE\)

From the given figure, we can also say that :

\(\angle FDC\)

\(\angle FDC\)

Using the above rule for \(\triangle CDF\)

\(\angle CDF\)

\(90Â°\)

\(\angle DCF\)

Also,

\(\angle DCF\)

\(30Â°\)

Or,

x = \(180Â°\)

*11. In Fig, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If \(\angle AFE=130Â°\), find*

*(i) \(\angle BDE\)*

*(ii) \(\angle BCA\)*

*(iii) \(\angle ABC\)*

(i)

Here,

\(\angle BAF\)

\(\angle FAD\)

Also,

\(\angle AFE\)

\(\angle ADF\)

\(\angle ADF\)

(ii) We know that the sum of all the angles of a triangle is \(180Â°\)

Therefore, for \(\triangle BDE\)

\(\angle BDE\)

\(\angle DBE\)

Also,

\(\angle FAD\)

\(90Â°\)

Or,

\(\angle ACB\)

(iii) \(\angle ABC\)

*12. ABC is a triangle in which \(\angle B=\angle C\) and ray AX bisects the exterior angle DAC. If \(\angle DAX=70Â°\). Find \(\angle ACB\). *

Here,

\(\angle CAX\)

\(\angle CAX\)

\(\angle CAX\)

\(70Â°\)

\(\angle CAB\)

\(\angle CAB\)

\(\angle ACB\)

\(\angle ACB\)

2\(\angle ACB\)

\(\angle ACB\)

\(\angle ACB\)

*13. The side BC of \(\triangle ABC\) is produced to a point D. The bisector of \(\angle A\) meets side BC in L. If \(\angle ABC=30Â°\) and \(\angle ACD=115Â°\), find \(\angle ALC\) *

\(\angle ACD\)

\(\angle ACD\)

\(115Â°\)

\(\angle ACB\)

\(\angle ACB\)

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

Therefore, for \(\triangle ABC\)

\(\angle ABC\)

\(30Â°\)

Or,

\(\angle BAC\)

\(\angle LAC\)

Using the above rule for \(\triangle ALC\)

\(\angle ALC\)

\(\angle ALC\)

Or,

\(\angle ALC\)

\(\angle ALC\)

*14. D is a point on the side BC of \(\triangle ABC\). A line PDQ through D, meets side AC in P and AB produced at Q. If \(\angle A=80Â°\), \(\angle ABC=60Â°\) and \(\angle PDC=15Â°\), find *

*(i) \(\angle AQD\) *

*(ii) \(\angle APD\) *

\(\angle ABD\)

\(\angle ABC\)

\(60Â°\)

\(\angle QBC\)

\(\angle PDC\)

\(\angle BDQ\)

In \(\triangle QBD\)

\(\angle QBD\)

\(120Â°\)

\(\angle BQD\)

\(\angle BQD=45Â°\)

\(\angle AQD\)

In \(\triangle AQP\)

\(\angle QAP\)

\(80Â°\)

\(\angle APQ\)

\(\angle APD\)

*15. Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y *

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

(I) From the given figure, we can see that:

\(\angle ACB\)

\(75Â°\)

Or,

x = \(105Â°\)

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

Therefore, for \(\triangle ABC\)

\(\angle BAC\)

\(40Â°\)

Or,

y = \(65Â°\)

(ii)

x +\(80Â°\)

= x = \(100Â°\)

In \(\triangle ABC\)

x+ y+ \(30Â°\)

\(100Â°\)

= y = \(50Â°\)

(iii)

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

Therefore, for \(\triangle ACD\)

\(30Â°\)

Or,

y = \(50Â°\)

\(\angle ACB\)

\(\angle ACB\)

Using the above rule for \(\triangle ACD\)

x+ \(45Â°\)

= x = \(55Â°\)

(iv)

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

Therefore, for \(\triangle DBC\)

\(30Â°\)

\(\angle DBC\)

x + \(\angle DBC\)

x = \(80Â°\)

And,

y = \(30Â°\)

*16. Compute the value of x in each of the following figures*

(i) From the given figure, we can say that :

\(\angle ACD\)

Or,

\(\angle ACB\)

We can also say that :

\(\angle BAE\)

Or,

\(\angle BAC\)

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

Therefore, for \(\triangle ABC\)

x+ \(\angle BAC\)

x = \(180Â°\)

= x = \(52Â°\)

(ii) From the given figure, we can say that :

\(\angle ABC\)

\(\angle ABC\)

We can also say that :

\(\angle ACB\)

\(\angle ACB\)

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

Therefore, for \(\triangle ABC\)

x+ \(\angle ABC\)

= x = \(50Â°\)

(iii)

From the given figure, we can see that :

\(\angle BAD\)

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

Therefore, for \(\triangle DEC\)

x + \(40Â°\)

= x = \(88Â°\)

(iv) In the given figure, we have a quadrilateral whose sum of all angles is \(360Â°\)

Thus,

\(35Â°\)

Or,

reflex\(\angle ADC\)

\(230Â°\)

= x = \(130Â°\)