In this page, you can find the complete solutions of the exercise of Properties of Triangle chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.


In the above-mentioned book, Solutions of Triangle is the 7th chapter and has only one exercise. This is the solution of that exercise.


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.


1 a) $a (b cos C \ - \ c cos B) \ = \ b^2 \ - \ c^2$

Solution:

LHS

$= \ a(b cosC \ - \ c cosB)$

$= \ abcosC \ - \ accosB)$

Using cosine law

$= \ \dfrac{a^2 + b^2 - c^2}{2}  \ - \ \dfrac{a^2 + c^2 - b^2}{2}$

$= \ \dfrac{a^2 + b^2 - c^2 -a^2 - c^2 + b^2}{2}$

$= \ \dfrac{2(b^2 \ - \ c^2)}{2}$

$= \ b^2 \ - \ c^2$

RHS


1 b) $\dfrac{cosA}{a} \ + \ \dfrac{a}{bc}$ = $\dfrac{cosB}{b} \ + \ \dfrac{b}{ca}$ = $\dfrac{cosC}{c} \ + \ \dfrac{c}{ab}$

Solution:

LHS

$= \ \dfrac{cosA}{a} \ + \ \dfrac{a}{bc}$

$= \ \dfrac{bc \ cosA + a^2}{abc}$

$= \ \dfrac{\frac{b^2 + c^2 - a^2}{2} + a^2}{abc}$

$= \ \dfrac{ b^2 + c^2 - a^2 + 2a^2}{2abc}$

$= \ \dfrac{a^2 + b^2 +c^2}{2abc}$

Similarly,

MS = $\dfrac{a^2 + b^2 + c^2}{2}$ = RHS


1c) $\dfrac{cosA}{a} \ + \ \dfrac{cosB}{b} \ + \ \dfrac{cosC}{c} \ = \ \dfrac{a^2 + b^2 + c^2}{2abc}$

Solution:

LHS

$= \ \dfrac{cosA}{a} \ + \ \dfrac{cosB}{b} \ + \ \dfrac{cosC}{c}$

$= \ \dfrac{bc cosA \ + \ ca cosB \ + \ ab cosC}{abc}$

Using cosine law

$= \ \dfrac{ (b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)}{2abc}$

$= \ \dfrac{a^2 + b^2 + c^2}{2abc}$

RHS


























5 a) $\dfrac{b-c}{a} cos \frac{1}{2} A \ = \ sin \frac{1}{2} (B-C)$

Solution:

LHS

$= \ \dfrac{b-c}{a} cos \frac{1}{2}A$

$= \ \dfrac{2RsinB \ - \ 2RsinC}{2RsinA} \cos \frac{1}{2} A$

$= \ \dfrac{sin B \ - \ sin C}{sinA} \cos \frac{1}{2} A$

$= \ \dfrac{ 2 \cos ( \frac{B+C}{2} ) \ .  \ \sin ( \frac{B-C}{2} )}{2 \sin \frac{1}{2} A \cos \frac{1}{2} A} \ cos \frac{1}{2}A$

$= \ \dfrac{\sin \frac{1}{2} A \ . \sin \frac{B-C}{2} }{ \sin \frac{1}{2} A}$

$= \ \cos \frac{1}{2} A$

RHS














6) If $a^4 + b^4 + c^4 = 2c^2 (a^2 + b^2)$, prove that C = 45$^o$ or 135$^o$.

Solution:

Given,

$a^4 + b^4 + c^4 = 2c^2 (a^2 + b^2)$

$or, \ (a^2 + b^2)^2 - 2a^2b^2 - 2c^2(a^2 +b^2) + c^4 = 0$

$or, \ (a^2 + b^2)^2 - 2(a^2+b^2) + (c^2)^2 = 2 a^2b^2$

$or, \ (a^2 + b^2 - c^2)^2 = 2a^2b^2$

$or, \ a^2 + b^2 - c^2 = \pm \sqrt{2} ab$

Using cosine law for angle C

$or, \ 2 ab cos C = \pm \sqrt{2} ab$

$or, \sqrt{2} cos C = \pm 1$

$or, \ cos C = \pm \dfrac{1}{\sqrt{2}}$

Either

$cos C = \dfrac{1}{\sqrt{2}}$

$\implies C = 45^o$

Or

$cosC = - \dfrac{1}{\sqrt{2}}$

$\implies C = 135^o$

7) If $(a+b+c)(b+c-a) = 3bc$, show that A = 60$^o$.

Solution:

Given,

$(a + b + c)(b + c- a) = 3bc$

$or, \{ (b+c) + a \} \{(b + c) - a \} = 3bc$

$or, (b+c)^2  - a^2 = 3bc$

$or, b^2 + c^2 - a^2 + 2bc = 3bc$

Using consine law for angle A

$or, 2 bc cos A = bc$

$or, cosA = \dfrac{1}{2}$

$or, cos A = cos 60^o$

$\therefore A = 60^o$

8) If $\dfrac{1}{a+c} + \dfrac{1}{b+c} = \dfrac{3}{a+b+c}$, show that C = 60$^o$.

Solution:

Given,

$\dfrac{1}{a+c} + \dfrac{1}{b+c} = \dfrac{3}{a+b+c}$

$or, \ \dfrac{b + c + a +c}{(a+c)(b+c)} = \dfrac{3}{a+b+c}$

$or, \ (a+b+c) \{ (a+b+c) + c \} = 3 (a +c)(b+c)$

$or,  \ (a+b+c)^2 + c(a+b+c) = 3 (ab + ac + bc + c^2)$

$or, \ a^2 + 2ab + b^2 + c^2 + 2ac + 2bc + ac + bc + c^2 = 3ab + 3ac + 3bc + 3c^2$

$or, \ a^2 + 2ab + b^2 + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc +3c^2$

$or, \ a^2 + 2ab + b^2 - c^2 = 3ab$

$or, \ a^2 + b^2 - c^2 = ab$

$or, \ 2ab \cos C = ab$

$or, \ \cos C = \dfrac{ab}{2ab}$

$or, \ \cos C = \dfrac{1}{2}$

$or, \ \cos C = \cos 60^o$

$\therefore C = 60^o$

9) If the cosines of two of the angles of a triangle are proportional to the opposite sides, prove that the triangle is isosceles.

Solution:

According to the question, in any triangle ABC,

$$\dfrac{\cos B}{\cos C} = \dfrac{\text{b}}{\text{c}}$$

$$\text{or,} \dfrac{\cos B}{\cos C} = \dfrac{2R \sin B }{ 2R \sin C}$$

$$\text{or,} \dfrac{\sin C}{\cos C} = \dfrac{\sin B}{\sin C}$$

$$\text{or,} \tan C = \tan B$$

$$\therefore \ C \ = \ B$$

Hence, the given triangle is isoceles triangle.

10. If (cos A + 2 cos C): (cos A + 2 cos B) = sinB:sinC, prove that the triangle is either isosceles or right angled.

Solution:

Given,

$$\dfrac{\cos A + 2 \cos C}{\cos A + 2 \cos B} = \dfrac{\sin B}{\sin C}$$

$\text{or,} \cos A \sin C + 2 \cos C \sin C = \cos A \sin B + 2 \sin B \cos B$

$\text{or,} \cos A \sin C - \cos A \sin B = \sin 2B - \sin 2C$

$\text{or,} \cos A ( \sin C - \sin B) \ = \ 2  \cos \frac{2B + 2C}{2} \sin \frac{2B - 2C}{2}$

$\text{or,} \cos A ( 2 \cos \frac{C+B}{2} \sin \frac{C - B}{2}) \ = \ 2 \cos (B+C) \sin (B-C)$

$\text{or,} \cos A \cos \frac{B+C}{2}  \sin \frac{C-B}{2} = - \cos A \sin (B-C)$

$\text{or,} \cos A \cos \frac{B+C}{2} \sin \frac{C-B}{2} =  \cos A \sin (C-B)$

$\text{or,} \cos A \cos \frac{B+C}{2} \sin \frac{C-B}{2} = \cos A 2 \sin \frac{C-B}{2} \cos \frac{C-B}{2}$

$\text{or,} \cos A \cos \frac{B+C}{2} \sin \frac{C-B}{2} - 2 \cos A \sin \frac{C-B}{2} \cos \frac{C-B}{2} = 0$ 

$\text{or,} \cos A \sin \frac{C-B}{2} \left ( \cos \frac{B+C}{2} - 2 \frac{C-B}{2} \right ) = 0$

Either

$ \cos A = 0$

$\therefore A = 90^o$

Hence, the triangle is right angled triangle.

Or

$ \sin \frac{C-B}{2} = 0$

$\text{or,} \sin \frac{C-B}{2} = sin 0$

$\text{or,} \frac{C-B}{2} = 0$

$\text{or,} C - B = 0$

$\therefore C = B$

Hence, the triangle is isoceles triangle.

But

 $\cos \frac{B+C}{2} - 2 \frac{C-B}{2} \neq 0$

11. If 2 cos A = sinB:sinC, show that the triangle is isosceles.

Solution:

Given,

$$2 \cos A = \dfrac{ \sin B}{\sin C}$$

$$\text{or,} \dfrac{b^2 + c^2 - a^2}{bc}\  = \ \dfrac{b}{c}$$

$$\text{or,} \dfrac{b^2 + c^2 - a^2}{b} \ = \ b$$

$$\text{or,} b^2 + c^2 - a^2 \ = \ b^2$$

$$\text{or,} c^2 \  = \ a^2$$

$$ c \ = \ a$$

Hence, the given triangle is isoceles triangle.

12. Prove that a^2, b^2, c^2 are in A.P. if sinA:sinC = sin(A-B):sin(B-C).

Solution:

Given,

$$\dfrac{\sin A}{\sin C} = \dfrac{\sin (A-B) }{\sin (B-C)}$$

$$\text{or,} \sin A \sin (B-C) \ = \ \sin C \sin (A-B)$$

In any triangle, A + B + C = $\pi$

$$\text{or,} \sin (B+C)  \sin (B-C) \ = \ \sin(A+B) \sin (A-B)$$

$$\text{or,}  \sin^2 B - \sin^2 C \ = \ \sin^2 A - \sin^2 B$$

Using Extended sine law

$$\text{or,} \dfrac{b^2}{4R^2} - \dfrac{c^2}{4R^2} \ = \ \dfrac{a^2}{4R^2} - \dfrac{b^2}{4R^2}$$

$$\text{or,} b^2 - c^2 \ = \ a^2 - b^2$$

$$\text{or,} 2b^2 \ = \ a^2 + c^2$$

$$ \therefore b^2  \ = \ \dfrac{a^2 + c^2}{2}$$







13 c) $\sin A + \sin B + \sin C = \dfrac{s}{R}$

Solution:

We know,

$$\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$$

$$= 2 \sqrt{ \dfrac{(s-b)(s-c)}{bc} } \sqrt{ \dfrac{s(s-a)}{bc}}$$

$$= 2 \sqrt{ \dfrac{(s(s-a)(s-b)(s-c)}{(bc)^2}}$$

$$= \dfrac{2 \triangle}{bc}$$

Similarly,

LHS

$$\sin A + \sin B + \sin C$$

$$\dfrac{a}{2R} + \dfrac{b}{2R} + \dfrac{c}{2R}$$

$$\dfrac{1}{2R} (a+b+c)$$

$$\dfrac{1}{2R} (2s)$$

$$\dfrac{s}{R}$$

RHS


13 d) In any triangle, prove that: acosBcosC + bcosCcosA + ccosAcosB = △/R

Solution:

To prove: $a \ cos B \ cos C \ + \ b \ cosC \ cosA \ + \ c \ cosA \ cos B \ = \ \dfrac{\triangle}{R}$

LHS

$= a \ cos B \ cos C \ + \ b \ cosC \ cosA \ + \ c \ cosA \ cos B$

$= cos C \ ( a cos B \ + \ b cos A ) \ + \ c \ cosA \ cos B$

$= cos C \times c \ + \ c \ cosA \ cos B$

$= c \ ( cos C \ + \ cos A \ cos B)$

$= \dfrac{c}{2} \ (2cos C \ + \ 2 cos A \ cos B)$

$= \dfrac{c}{2} \ ( 2 cos C \ + \ cos(A + B) + cos (A-B) )$

$= \dfrac{c}{2} \ ( 2 cos C \ + \ cos( \pi - C) + cos (A -B) )$

$= \dfrac{c}{2} \ (2 cos C \ - \ cos C \ + \ cos (A -B))$

$= \dfrac{c}{2} \ ( cos C \ + \ cos (A-B) )$

$= \dfrac{c}{2} \ ( cos (\pi - (A+B)) \ + \ cos (A-B))$

$= \dfrac{c}{2} \ (cos (A-B) \ - \ cos(A+B))$

$= \dfrac{c}{2} ( 2 \ sin A \ sin B)$

$= sin A \ sin B \ c$

$= \dfrac{a}{2R} \ \times \ \dfrac{b}{2R} \ \times c$

$= \dfrac{abc}{4R^2}$

$= \dfrac{abc}{4R} \ \times \ \dfrac{1}{R}$

$= \dfrac{ \triangle}{R}$

RHS

15) If a = 10, b = 8, c = 6, find s, $|traingle$, R and sin $\frac{B}{2}$.

Solution:

Given,

$a = 10$

$b = 8$

$c = 6$

$s = \dfrac{a+b+c}{2} = \dfrac{10+8+6}{2}$

$\therefore s = 12$

We know,

$\triangle = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{12 ( 12 - 10)(12- 8)(12-6)}$

$= \sqrt{12 (2)(4)(6)}$

$= \sqrt{576}$

$= 24$

And,

$\triangle = \dfrac{abc}{4R}$

$\implies R = \dfrac{abc}{4 \triangle}$

$= \dfrac{10 \times 8 \times 6}{4 \times 24}$

$= \dfrac{5}$

Also,

$\sin \frac{B}{2} = \sqrt{ \dfrac{(s-a)(s-c)}{ac}}$

$= \sqrt{ \dfrac{(12-10)(12-6)}{10 \times 6}}$

$= \sqrt{ \dfrac{2 \times 6}{60}}$

$= \sqrt{ \dfrac{1}{5}}$

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)