In this page, you can find the complete solutions of the third exercise of Review of Composition and Resolution of Vectors chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, circle is the 13th chapter and has three exercises only. Out of which, this is the solution of the third and final exercise.

Check: Grade 11 Circle Exercise 1 Complete Solutions

Grade 11 Circle Exercise 1 Solutions | Basic Mathematics Grade XI by Sukunda Pustak Bhawan

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.

6) If vec{a} and \vec{b} are two vectors of unit length and \theta is the angle between them, show that

\dfrac{1}{2} | \vec{a} \  - \ \vec{b} | \ = \ \sin \dfrac{\theta}{2}

Solution:

Given,

\text{A} = \frac{\pi}{2}

LHS

= \dfrac{1}{2} | \vec{a} \ - \ \vec{b} |

= \sqrt{ \left ( \dfrac{1}{2} | \vec{a} \ - \ \vec{b} |  \right )^2}

= \sqrt{ \dfrac{1}{4} \left ( a^2 \ + \ b^2 \ + \ 2 \vec{a} \vec{b} \cos \theta  \right ) }

|a| \ =  \  |b| \ = \ 1

= \sqrt{ \dfrac{1}{4} \left ( 1 + 1 + 2 \cos \theta \right )}

= \sqrt{ \dfrac{1}{4} \times 2 ( 1 + \cos \theta)}

= \sqrt{ \dfrac{1}{2} \times 2 \sin^2 \dfrac{\theta}{2}}

= \sqrt{ \sin^2 \dfrac{\theta}{2} }

= \sin \dfrac{\theta}{2}

RHS

7) In a right angled triangle ABC, right angle at A, show that AB^2 + AC^2 = BC^2

Solution:

In a triangle, let AB = \vec{c}, AC = \vec{b} and BC = \vec{a}

We know,

\vec{a} + \vec{b} + \vec{c} = 0

\vec{a} = - ( \vec{b} + \vec{c} )

\text{Squaring both sides}

a^2 = b^ 2 + c^2 + 2bc \cos ( \pi - A)

a^2 = b^2 + c^2 + 2bc \cos ( \pi - \frac{\pi}{2} )

a^2 = b^2 + c^ 2 + 2bc \cos \frac{\pi}{2}

a^2 = b^2 + c^2 + 0

a^2 = b^2 + c^2

\therefore \text{BC}^2 = \text{AC}^2 + \text{AB}^2

\therefore \text{AB}^2 + \text{AC}^2 = \text{BC}^2

9. Prove vectorically that

a) (i) b = c cosA + a cos C

Solution:

In a triangle, let AB = \vec{c}, AC = \vec{b} and BC = \vec{a}, taken in order.

We know,

\vec{a} + \vec{b} + \vec{c} = 0

\vec{b} = - ( \vec{a} + \vec{c} )

\text{Multiply both sides scalarly by } \vec{b}

\vec{b}.\vec{b} = - ( \vec{a} + \vec{c} ) \vec{b}

b^2 = - \vec{a}.\vec{b} - \vec{c}.\vec{b}

b^2 = -  ab \cos ( \pi - C) -  cb \cos ( \pi - A)

b^2 = -  ab (- \cos c) - cb (- \cos A)

b^2 = ab \cos C + bc \cos A

b = a \cos C + c \cos A

\therefore b = c \cos A + a \cos C

a) (ii) c = a cosB + b cos A

Solution:

In a triangle, let AB = \vec{c}, AC = \vec{b} and BC = \vec{a}, taken in order.

We know,

\vec{a} + \vec{b} + \vec{c} = 0

\vec{c} = - ( \vec{a} + \vec{b} )

\text{Multiply both sides scalarly by } \vec{c}

\vec{c}.\vec{c} = - ( \vec{a} + \vec{b} ) \vec{c}

c^2 = - \vec{a}.\vec{c} - \vec{b}.\vec{c}

c^2 = -  ac \cos ( \pi - B) -  bc \cos ( \pi - A)

c^2 = -  ac (- \cos B) - cb (- \cos A)

c^2 = ac \cos B + bc \cos A

c = a \cos B + b \cos A

\therefore c = b \cos A + a \cos B

About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)