Solution of Triangle Exercise: 8.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Solution of Triangle

Exercise 7.1

1. (i) The angles of a triangle are 105° and 15°. Find the ratio of its sides.

Solution:

In $\Delta $ABCm Let A = 105°, B = 15° Then C = 180 – (A + B) = 180 – (105+15) = 60°.

Now, we have,

Or, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{{\rm{sin}}105^{\circ} }} = \frac{{\rm{b}}}{{{\rm{sin}}15^{\circ} }} = \frac{{\rm{c}}}{{{\rm{sin}}60^{\circ} }}$

Or, $\frac{{\rm{a}}}{{\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}}}$ = $\frac{{\rm{b}}}{{\frac{{\sqrt 3  - 1}}{{2\sqrt 2 }}}}$ =

Or, $\frac{{\rm{a}}}{{\sqrt 3  + 1}}$ = $\frac{{\rm{b}}}{{\sqrt 3  - 1}}$ = $\frac{{\rm{c}}}{{\sqrt 6 }}$

So, a : b : c = $\left( {\sqrt 3  + 1} \right):\left( {\sqrt 3  - 1} \right):\sqrt 6 $

 

(ii). If A = 45°, B = 60°, show that a:c = 2 : $\left( {\sqrt 3  + 1} \right)$

Solution:

Here, A = 45°, B = 60°.

So, C = 180 – (A + B) = 180 – (45 + 60) = 75°.

To show: a : c = 2 : $\left( {\sqrt 3  + 1} \right)$

Since, we have, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{{\rm{sin}}45^{\circ} }}$ = $\frac{{\rm{c}}}{{{\rm{sin}}75^{\circ} }}$

Or, $\frac{{\rm{a}}}{{\frac{1}{{\sqrt 2 }}}}$ = $\frac{{\rm{c}}}{{\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}}}$

Or, $\frac{{\rm{a}}}{2}$ = $\frac{{\rm{c}}}{{\sqrt 3  + 1}}$

So, a : c = 2 : $\left( {\sqrt 3  + 1} \right)$

 

(iii) If one angle of the triangle is 60° and the ratio of the other two is 1:3, find all the angles and the ratio of the sides.

Solution:

Here, In $\Delta $ABC,

A = 60° and B:C = 1:3

So, B = K, C = 3K.

We have, A + B + C = 180

Or, 60 + K + 3K = 180

Or, 4K = 120°

So, k = 30°

So, A = 60°

B = 30°

C = 90°.

And, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\rm{b}}}{{\frac{1}{2}}} = \frac{{\rm{c}}}{1}$

Or, $\frac{{\rm{a}}}{{\sqrt 3 }} = \frac{{\rm{b}}}{1} = \frac{{\rm{c}}}{2}$

Hence, a : b : c = $\sqrt 3 $ : 1 : 2 and A = 60°, B = 30°, C = 90°.

 

(iv).The angles of a triangle are in the ratio 2:3:7. Prove that the sides are in the ratio of $\sqrt 2 $ : 2 : $\left( {\sqrt 3  + 1} \right)$.

Solution:

Here in ∆ABC,

A : B : C = 2 : 3 : 7

A = 2K, B = 3K and C = 7K

We have, A + B + C = 180

Or, 2K + 3K + 7K = 180

So, k = 15°

So, A = 30°

B = 45°

C = 105°.

And, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{{\rm{sin}}30^{\circ} }} = \frac{{\rm{b}}}{{{\rm{sin}}45^{\circ} }} = \frac{{\rm{c}}}{{{\rm{sin}}105^{\circ} }}$

Or, $\frac{{\rm{a}}}{{\frac{1}{2}}} = \frac{{\rm{b}}}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}}}$

Or, $\frac{{\rm{a}}}{{\sqrt 2 }} = \frac{{\rm{b}}}{2} = \frac{{\rm{c}}}{{\sqrt 3  + 1}}$

Hence, a : b : c = $\sqrt 2 $ : 2 : $\left( {\sqrt 3  + 1} \right)$

 

(v). If cos A = $\frac{4}{5}$, cosB = $\frac{3}{5}$, find a:b:c.

Solution:

Here, cos A = $\frac{4}{5}$, cosB = $\frac{3}{5}$

So, sinA = $\frac{3}{5}$, sin B = $\frac{4}{5}$.

To find: a : b : c.

Now, sinC = sin(A + B)                    [A + B + C = 180°]

= sinA.cosB + cosA.sinB = $\frac{3}{5}{\rm{*}}\frac{3}{5} + \frac{4}{5}{\rm{*}}\frac{4}{5} = \frac{9}{{25}} + \frac{{16}}{2}$ = 1

So, sinC = 1

SO, C = 90°.

We have, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{\frac{3}{5}}} = \frac{{\rm{b}}}{{\frac{4}{5}}} = \frac{{\rm{c}}}{1}$

Or, $\frac{{\rm{a}}}{3} = \frac{{\rm{b}}}{4} = \frac{{\rm{c}}}{5}{\rm{\: }}$

So, a : b : c = 3 : 4 : 5.

 

2. (i). Given a = 2, b = $\sqrt 2 $, c =$\sqrt 3 $ + 1, solve the triangle.

Solution:

Given, a = 2, b = $\sqrt 2 $, c =$\sqrt 3 $ + 1

To find: A, B, C.

Now,

cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = $\frac{{2 + {{\left( {\sqrt 3  + 1} \right)}^2} - 4}}{{2.\sqrt 2 .\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2 + 3 + 2\sqrt 3  + 1 - 4}}{{2.\sqrt 2 .\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2 + 2\sqrt 3 }}{{2.\sqrt 2 \left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2\left( {1 + \sqrt 3 } \right)}}{{2.\sqrt 2 .\left( {\sqrt 3  + 1} \right)}}$

= $\frac{1}{{\sqrt 2 }}$ = cos45°.

Again.

Cos B = $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$ = $\frac{{{{\left( {\sqrt 3  + 1} \right)}^2} + 4 - 2}}{{2.\left( {\sqrt 3  + 1} \right).2}}$

= $\frac{{3 + 2\sqrt 3  + 1 + 4 - 2}}{{4.\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{6 + 2\sqrt 3 }}{{4.\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2\sqrt 3 \left( {\sqrt 3  + 1} \right)}}{{4\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{\sqrt 3 }}{2}$ = cos30°.

 

(ii). If a = 3 + $\sqrt 3 $, b = $2\sqrt 3 $, c =$\sqrt 6 $, solve the triangle.

Solution:

Given, a = 3 + $\sqrt 3 $, b = $2\sqrt 3 $, c =$\sqrt 6 $

To find: A, B, C.

Now,

cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = $\frac{{{{\left( {2\sqrt 3 } \right)}^2} + {{\left( {\sqrt 6 } \right)}^2} - {{\left( {3 + \sqrt 3 } \right)}^2}}}{{2.2.\sqrt 3 ..\sqrt 6 }}$

= $\frac{{12 + 6 - 9 - 6\sqrt 3  - 3}}{{4{\rm{*}}\sqrt {18} }}$

= $\frac{{6 - 6\sqrt 3 }}{{4{\rm{*}}3\sqrt 2 }}$

= $\frac{{6\left( {1 - \sqrt 3 } \right)}}{{12\sqrt 2 }}$

= $\frac{{ - \left( {\sqrt 3  - 1} \right)}}{{2\sqrt 2 }}$ = cos105°.

So, A = 105°.

Again.

Cos B = $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$ = $\frac{{6 + \left( {9 + 6\sqrt 3  + 3} \right) - 12}}{{2.\sqrt 6 .\left( {3 + \sqrt 3 } \right)}}$

= $\frac{{6 + 6\sqrt 3 }}{{2\sqrt 6 \left( {3 + \sqrt 3 } \right)}}$

= $\frac{{6(1 + \sqrt {3)} }}{{2.\sqrt 6 .\sqrt 3 .\left( {1 + \sqrt 3 } \right)}}$

= $\frac{3}{{\sqrt {18} }}$

= $\frac{3}{{3\sqrt 2 }}$

= $\frac{1}{{\sqrt 2 }}$ = cos45°.

So, B = 45°.

And C = 180 – (A + B) = 180 – (105 + 45) = 30°

Hence, A = 105°, B = 45°, C = 30°.

 

(iii) If three sides of the triangle are proportional to 2 : $\sqrt 6 $ : $\sqrt 3 $ + 1 find all the angles.

Solution:

Here, In $\Delta $ABC, we have,

a : b : c = 2 : $\sqrt 6 $ : $\sqrt 3 $ + 1

So, a = 2K, b = $\sqrt 6 $K, c = $\left( {\sqrt 3  + 1} \right){\rm{K}}$

To find, A, B, C.

Now, cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = $\frac{{6{{\rm{K}}^2} + \left( {3 + 2\sqrt 3  + 1} \right){{\rm{K}}^2} - 4{{\rm{K}}^2}}}{{2.\sqrt 6 {\rm{K}}.\left( {\sqrt 3  + 1} \right){\rm{K}}}}$

= $\frac{{6 + 3 + 2\sqrt 3  + 1 - 4}}{{2\sqrt 6 .\left( {\sqrt 3  + 1} \right)}}$ = $\frac{{6 + 2\sqrt 3 }}{{2.\sqrt 6 .\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2\sqrt 3 \left( {\sqrt 3  + 1} \right)}}{{2\sqrt 6 .\left( {\sqrt 3  + 1} \right)}}$ = $\frac{1}{{\sqrt 2 }}$ = cos 45°.

So, A = 45°.

Again.

Cos B = $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$ = $\frac{{\left( {3 + 2\sqrt 3  + 1} \right){{\rm{K}}^2} + 4{{\rm{K}}^2} - 6{{\rm{K}}^2}}}{{2.\left( {\sqrt 3  + 1} \right){\rm{K}}.2{\rm{K}}}}$

= $\frac{{3 + 2\sqrt 3  + 1 + 4 - 6}}{{4\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2 + 2\sqrt 3 }}{{4\left( {\sqrt 3  + 1} \right)}}$

= $\frac{{2\left( {1 + \sqrt 3 } \right)}}{{4\left( {1 + \sqrt 3 } \right)}}$

= $\frac{1}{2}$ = cos60°.

So, B = 60°.

And C = 180 – (A + B) = 180 – (45 + 60) = 75°

Hence, A = 45°, B = 60°, C = 75°.

 

3. (i) If C = 30°, B = 45°, C = 6$\sqrt 2 $. Solve the triangle.

Solution: C = 30°, B = 45°, C = 6$\sqrt 2 $

To find: a, b, A

Now, A = 180 – (B + C) = 180 – (45 + 30) = 105°.

And, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{{\rm{sin}}105^{\circ} }} = \frac{{\rm{b}}}{{{\rm{sin}}45^{\circ} }} = \frac{{\rm{c}}}{{{\rm{sin}}30^{\circ} }}$

Or, $\frac{{\rm{a}}}{{\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}}} = \frac{{\rm{b}}}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{1}{2}}}$

Or, $\frac{{\rm{a}}}{{\sqrt 3  + 1}} = \frac{{\rm{b}}}{2} = \frac{{6\sqrt 2 }}{{\sqrt 2 }}$   [c = 6$\sqrt 2 $]

So, a = $\frac{{6\sqrt 2 }}{{\sqrt 2 }}\left( {\sqrt 3  + 1} \right)$ = 6($\sqrt 3 $ + 1) and b = $\frac{{6\sqrt 2 }}{{\sqrt 2 }}.2$ = 12.

Hence, A = 105°, a = 6($\sqrt 3 $ + 1), b = 12.

 

(ii) If A = 60°, B = 75°, C = 2$\sqrt 3 $, solve the triangle.

Solution: A = 60°, B = 75°, C = 2$\sqrt 3 $

To find: b, c, C

Now, C = 180 – (A + B) = 180 – (60 + 75) = 45°.

And, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{2\sqrt 3 }}{{{\rm{sin}}60^{\circ} }} = \frac{{\rm{b}}}{{{\rm{sin}}75^{\circ} }} = \frac{{\rm{c}}}{{{\rm{sin}}45^{\circ} }}$

Or, $\frac{{2\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\rm{b}}}{{\frac{{\sqrt 3  + 1}}{{2.\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{1}{{\sqrt 2 }}}}$

Or, $\frac{{2\sqrt 3 }}{{\sqrt 3 .\sqrt 2 }} = \frac{{\rm{b}}}{{\sqrt 3  + 1}} = \frac{{\rm{c}}}{2}$

Or, $\sqrt 2 $ = $\frac{{\rm{b}}}{{\sqrt 3  + 1}}$ = $\frac{{\rm{c}}}{2}$.

So, C = 45°, b = $\sqrt 6  + \sqrt 2 $, c = 2$\sqrt 2 $.

 

(iii) If A = 30°, B = 45°, b = 2.Solve the triangle.

Solution: A = 30°, B = 45°, b = 2

To find: a, c, C

Now, C = 180 – (A + B) = 180 – (30 + 45) = 105°.

And, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\rm{a}}}{{{\rm{sin}}30^{\circ} }} = \frac{{\rm{b}}}{{{\rm{sin}}45^{\circ} }} = \frac{{\rm{c}}}{{{\rm{sin}}105^{\circ} }}$

Or, $\frac{{\rm{a}}}{{\frac{1}{2}}} = \frac{2}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{{\sqrt 3  + 1}}{{2.\sqrt 2 }}}}$

Or, $\frac{{\rm{a}}}{{\sqrt 2 }} = \frac{2}{2} = \frac{{\rm{c}}}{{\sqrt 3  + 1}}$

Or, $\frac{{\rm{a}}}{{\sqrt 2 }}$ = $1$ = $\frac{{\rm{c}}}{{\sqrt 3  + 1}}$.

So, a = $\sqrt 2 $, c = $\sqrt 3  + 1$.

Hence, C = 105°, a = $\sqrt 2 $, c = $\sqrt 3 $ + 1.

 

4. (i) If a = 2, b = 4, c = 60°.Find A and B.

Solution:

Here, a = 2, b = 4, c = 60°.

To find: A and B.

We have, cosC = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$

Or, $\frac{1}{2}$ = $\frac{{4 + 16 - {{\rm{c}}^2}}}{{2.2.4}}$    [ C = 60°]

Or, 8 = 20 – c².

Or, c² = 12

So, c = 2$\sqrt 3 $.

Now, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{2}{{{\rm{sinA}}}} = \frac{4}{{{\rm{sinB}}}} = \frac{{2\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}}}$      [c = $2\sqrt 3 $ and C = 60°]

Or, $\frac{2}{{{\rm{sinA}}}}$ = $\frac{4}{{{\rm{sinB}}}}$ = 4

Now, sinA = $\frac{1}{2}$à A = 30° and sinB = 1 à B = 90°.

Hence, A = 30°, B = 90°.

 

(ii) Two sides of triangle are $\sqrt 3 $ _- 1 and $\sqrt 3 $ - 1 and the included angle is 60°, solve the triangle.

Solution:

Here, b = $\sqrt 3 $ - 1, c = $\sqrt 3 $ + 1, A = 60°

To find : a, B, C.

We have, cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$

Or, $\frac{1}{2}$ = $\frac{{\left( {3 - 2\sqrt 3  + 1} \right) + \left( {3 + 2\sqrt 3  + 1} \right) - {{\rm{a}}^2}}}{{2.\left( {\sqrt 3  - 1{\rm{\: }}} \right)\left( {\sqrt 3  + 1} \right)}}$

Or, 3 – 1 = 8 – a².

Or, a² = 6.

So, a = $\sqrt 6 $.

Now, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{{\sqrt 6 }}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\sqrt 3  - 1{\rm{\: }}}}{{{\rm{sinB}}}} = \frac{{\sqrt 3  + 1{\rm{\: }}}}{{{\rm{sinC}}.}}$

Or, $2\sqrt 2 $ = $\frac{{\sqrt 3  - 1{\rm{\: }}}}{{{\rm{sinB}}}}$ = $\frac{{\sqrt 3  + 1{\rm{\: }}}}{{{\rm{sinC}}}}$

So, sinB = $\frac{{\sqrt 3  - 1{\rm{\: }}}}{{2.\sqrt 2 }}$ = sin15°   [165° is not possible]

So, B = 15°

And, sinC = $\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}$ = sin75° or sin105°.

So, C = 105°   [c = 75°, not possible in this case]

Hence, a = $\sqrt 6 $, B = 15°, C = 105°.

 

(iii) Given a = $\sqrt {57} $, A = 60° $\Delta $ = 2$\sqrt 3 $. Find b and c.

Solution:

Here, a = $\sqrt {57} $, $\Delta $ = 2$\sqrt 3 $, A = 60°

To find: b and c.

Since, $\Delta $ = $\frac{1}{2}$ bc.sinA.

Or, 2$\sqrt 3 $ = $\frac{1}{2}$ bc. $\frac{{\sqrt 3 }}{2}$    [sinA = sin60° = $\frac{{\sqrt 3 }}{2}$]

So, bc = 8 …(i)

Also, cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$

Or, $\frac{1}{2}$ = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - 57}}{{2.8}}$    [bc = 8]

Or, 8 = b² + c² – 57.

Or, b² + c² – 65 = 0

Or, b² + $\frac{{64}}{{{{\rm{b}}^2}}}$ - 65 = 0    [c = $\frac{8}{{\rm{b}}}$]

Or, b⁴ – 65b² + 64 = 0

Or. b⁴– 64b² – b² + 64 = 0

Or, b²(b² – 64) – 1(b² – 64) = 0

Or, (b² – 64)(b² – 1) =0

So, b = 1, 8.

Or, c = 8.1 [from(i)]

Hence, b = 1 or 8 and c = 8 or 1.

 

5. Given:

(i) a = 3, b = 3$\sqrt 3 $, A = 30°

Solution:

Here, a = 3, b = 3$\sqrt 3 $, A = 30°

To find: c, B, C.

Now, $\frac{{\rm{a}}}{{{\rm{sinA}}}}$ = $\frac{{\rm{b}}}{{{\rm{sinB}}}}$

Or, $\frac{3}{{{\rm{sin}}30^{\circ} }} = \frac{{3\sqrt 3 }}{{{\rm{sinB}}}}$

Or, sinB = $\frac{{\sqrt 3 }}{2}$    [sin30 = $\frac{1}{2}$]

So, B = 60° or 120°. (Ambiguous case).

Now,

a. When B = 60°.

C = 180 – (A + B) = 180 – (30 + 60) = 90°

Again, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{3}{{\frac{1}{2}}} = \frac{{\rm{c}}}{1}$à c = 6.

So, C = 90°, c = 6, B = 60°.

 

 

b. When B = 120°.

C = 180 – (A + B) = 180 – (30 + 120) = 30°

Again, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{3}{{\frac{1}{2}}} = \frac{{\rm{c}}}{{\frac{1}{2}}}$à c = 3.

So, B = 120°, C = 30°, c = 3.

Hence, the two solutions are, B = 60°, C = 90°, c = 6 and B =120°, C = 30°, c = 3.

 

(ii) a=2, b=$\sqrt 3 $ + 1, A = 45°.

Solution:

Here, a = 2, b = $\sqrt 3 $ + 1, A = 45°

To find: c, B, C.

Now, $\frac{{\rm{a}}}{{{\rm{sinA}}}}$ = $\frac{{\rm{b}}}{{{\rm{sinB}}}}$

Or, $\frac{2}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\sqrt 3  + 1}}{{{\rm{sinB}}}}$

Or, sinB = $\frac{{\sqrt 3  + 1}}{{2\sqrt 2 }}$ = sin75° or sin105° 

So, B = 75° or 105°. (Ambiguous case).

Now,

a.

When B = 75°.

C = 180 – (A + B) = 180 – (45 + 75) = 60°

Again, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{2}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{{\sqrt 3 }}{2}}}$à 2$\sqrt 2 $ = $\frac{{2{\rm{C}}}}{{\sqrt 3 }}$.

So, c = $\sqrt 6 $

So, B = 75°, C = 60°, c = $\sqrt 6 $.

 

b.

When B = 105°.

C = 180 – (A + B) = 180 – (45 + 105) = 30°

Again, $\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{2}{{\frac{1}{{\sqrt 2 }}}} = \frac{{\rm{c}}}{{\frac{1}{2}}}$à c = $\sqrt 2 $.

So, B = 105°, C = 30°, c = $\sqrt 2 $.

Hence, the two solutions are, B = 75°, C = 60°, c = $\sqrt 6 $ and B =105°, C = 30°, c = $\sqrt 2 $.

 

(iii) A = 30°, a = 6, b = 4 (Take 19°30’ = 1/3 and sin 49°30’ =3/4)

Solution:

Here, A = 30°, a = 6, b = 4

To find: B, C, c

Now, $\frac{{\rm{a}}}{{{\rm{sinA}}}}$ = $\frac{{\rm{b}}}{{{\rm{sinB}}}}$

Or, $\frac{6}{{\frac{1}{2}}} = \frac{4}{{{\rm{sinB}}}}$

Or, 12 = $\frac{4}{{{\rm{sinB}}}}$

Or, sinB = $\frac{1}{3}$ = sin19° 30’ (given)

So, B = 19° 30’. (No two solutions)

Now,

C = 180 – (A + B) = 180 – (30° + 19°30’) = 130°30’

Again. $\frac{{\rm{a}}}{{{\rm{sinA}}}}$ = $\frac{{\rm{c}}}{{{\rm{sinC}}}}$

Or, $\frac{6}{{\frac{1}{2}}}$ = $\frac{{\rm{c}}}{{{\rm{sin}}130^{\circ} {{30}^{\rm{'}}}}}$

Or, 12 = $\frac{{\rm{c}}}{{\frac{3}{4}}}$   [sin 130°30’ = sin 49°30’ = $\frac{3}{4}$] Given

Or, c = 9.

Hence, B = 19°30’, C = 130°30’, c = 9