📐 Chapter 7: Solution of Triangle | NEB Mathematics Notes Class 12
Complete Solutions with Sine Law, Cosine Law & Ambiguous Case | Based on Latest Syllabus
✅ Updated according to latest syllabus of 2080 | Complete exercise solutions included
🔺 Solution of Triangle — Solution of Triangle Class 12 Mathematics Notes has been updated according to the latest syllabus of 2080. A triangle has three sides and three angles — called the six elements of a triangle. To solve a triangle means to find the remaining elements when three other elements are known. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📖 Fundamental Formulas
Sine Law: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R\)
Cosine Law: \(\cos A = \dfrac{b^2 + c^2 – a^2}{2bc}\), \(\cos B = \dfrac{c^2 + a^2 – b^2}{2ca}\), \(\cos C = \dfrac{a^2 + b^2 – c^2}{2ab}\)
Projection Formula: \(a = b\cos C + c\cos B\), \(b = c\cos A + a\cos C\), \(c = a\cos B + b\cos A\)
Angle Sum: \(A + B + C = 180°\)
Circum-radius: \(R = \dfrac{abc}{4\Delta}\)
Cosine Law: \(\cos A = \dfrac{b^2 + c^2 – a^2}{2bc}\), \(\cos B = \dfrac{c^2 + a^2 – b^2}{2ca}\), \(\cos C = \dfrac{a^2 + b^2 – c^2}{2ab}\)
Projection Formula: \(a = b\cos C + c\cos B\), \(b = c\cos A + a\cos C\), \(c = a\cos B + b\cos A\)
Angle Sum: \(A + B + C = 180°\)
Circum-radius: \(R = \dfrac{abc}{4\Delta}\)
📌 Five Cases in Solving a Triangle:
1. Three angles given (only ratios of sides found)
2. Three sides given
3. Two angles and one side given
4. Two sides and the included angle given
5. Two sides and an angle opposite one of them given (Ambiguous case)
1. Three angles given (only ratios of sides found)
2. Three sides given
3. Two angles and one side given
4. Two sides and the included angle given
5. Two sides and an angle opposite one of them given (Ambiguous case)
📝 Exercise 7.1 — Solutions
Q1(i)
Find \(a:b:c\) given \(\angle B = 15°,\ \angle C = 105°\)
Given: \(B = 15°,\ C = 105°\)
Step 1 — Find A:
\(\angle BAC + \angle ACB + \angle CBA = 180°\quad[\because \text{sum of angles of a } \triangle]\)
\(\Rightarrow x + 15° + 105° = 180°\)
\(\Rightarrow x + 120° = 180°\)
\(\therefore A = 60°\)
Step 2 — Apply Sine Rule:
\(a : b : c = \sin A : \sin B : \sin C\)
\(= \sin 105° : \sin 15° : \sin 60°\)
\(= \dfrac{\sqrt{3}+1}{2\sqrt{2}} : \dfrac{\sqrt{3}-1}{2\sqrt{2}} : \dfrac{\sqrt{3}}{2}\)
\(= \sqrt{3}+1 : \sqrt{3}-1 : \sqrt{6}\)
Step 1 — Find A:
\(\angle BAC + \angle ACB + \angle CBA = 180°\quad[\because \text{sum of angles of a } \triangle]\)
\(\Rightarrow x + 15° + 105° = 180°\)
\(\Rightarrow x + 120° = 180°\)
\(\therefore A = 60°\)
Step 2 — Apply Sine Rule:
\(a : b : c = \sin A : \sin B : \sin C\)
\(= \sin 105° : \sin 15° : \sin 60°\)
\(= \dfrac{\sqrt{3}+1}{2\sqrt{2}} : \dfrac{\sqrt{3}-1}{2\sqrt{2}} : \dfrac{\sqrt{3}}{2}\)
\(= \sqrt{3}+1 : \sqrt{3}-1 : \sqrt{6}\)
✅ \(a:b:c = \sqrt{3}+1 : \sqrt{3}-1 : \sqrt{6}\)
Q1(ii)
Find \(a:c\) given \(A = 45°,\ B = 60°\)
Given: \(A = 45°,\ B = 60°\)
Step 1 — Find C:
\(A + B + C = 180°\Rightarrow 45° + 60° + C = 180°\)
\(\therefore C = 75°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow \dfrac{a}{c} = \dfrac{\sin A}{\sin C} = \dfrac{\sin 45°}{\sin 75°} = \dfrac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{2}{\sqrt{3}+1}\)
Step 1 — Find C:
\(A + B + C = 180°\Rightarrow 45° + 60° + C = 180°\)
\(\therefore C = 75°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow \dfrac{a}{c} = \dfrac{\sin A}{\sin C} = \dfrac{\sin 45°}{\sin 75°} = \dfrac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{2}{\sqrt{3}+1}\)
✅ \(a:c = 2 : \sqrt{3}+1\)
Q1(iii)
Find \(a:b:c\) given \(A = 60°,\ B:C = 1:3\)
Given: \(A = 60°,\ B:C = 1:3\)
Step 1 — Let \(B = k,\ C = 3k\)
\(A + B + C = 180°\Rightarrow 60° + k + 3k = 180°\)
\(\Rightarrow 4k = 120°\Rightarrow k = 30°\)
\(\therefore B = 30°,\ C = 90°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{\sqrt{3}}{2}} = \dfrac{b}{\frac{1}{2}} = \dfrac{c}{1}\)
\(\Rightarrow \dfrac{a}{\sqrt{3}} = \dfrac{b}{1} = \dfrac{c}{2}\)
Step 1 — Let \(B = k,\ C = 3k\)
\(A + B + C = 180°\Rightarrow 60° + k + 3k = 180°\)
\(\Rightarrow 4k = 120°\Rightarrow k = 30°\)
\(\therefore B = 30°,\ C = 90°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{\sqrt{3}}{2}} = \dfrac{b}{\frac{1}{2}} = \dfrac{c}{1}\)
\(\Rightarrow \dfrac{a}{\sqrt{3}} = \dfrac{b}{1} = \dfrac{c}{2}\)
✅ \(a:b:c = \sqrt{3}:1:2\) and \(A=60°,\ B=30°,\ C=90°\)
Q1(iv)
Prove \(a:b:c = \sqrt{2}:2:(\sqrt{3}+1)\) given \(A:B:C = 2:3:7\)
Given: \(A:B:C = 2:3:7\). To prove: \(a:b:c = \sqrt{2}:2:(\sqrt{3}+1)\)
Step 1 — Let \(A = 2k,\ B = 3k,\ C = 7k\)
\(2k + 3k + 7k = 180°\Rightarrow 12k = 180°\Rightarrow k = 15°\)
\(\therefore A = 30°,\ B = 45°,\ C = 105°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = \dfrac{b}{\frac{1}{\sqrt{2}}} = \dfrac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\)
\(\Rightarrow \dfrac{a}{\sqrt{2}} = \dfrac{b}{2\sqrt{2}} = \dfrac{c}{\sqrt{3}+1}\)
Step 1 — Let \(A = 2k,\ B = 3k,\ C = 7k\)
\(2k + 3k + 7k = 180°\Rightarrow 12k = 180°\Rightarrow k = 15°\)
\(\therefore A = 30°,\ B = 45°,\ C = 105°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = \dfrac{b}{\frac{1}{\sqrt{2}}} = \dfrac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\)
\(\Rightarrow \dfrac{a}{\sqrt{2}} = \dfrac{b}{2\sqrt{2}} = \dfrac{c}{\sqrt{3}+1}\)
✅ Hence \(a:b:c = \sqrt{2}:2\sqrt{2}:(\sqrt{3}+1)\) — Proved
Q1(v)
Find \(a:b:c\) given \(\cos A = \dfrac{4}{5},\ \cos B = \dfrac{3}{5}\)
Given: \(\cos A = \dfrac{4}{5},\ \cos B = \dfrac{3}{5}\)
Step 1 — Find sin A and sin B:
Since \(\cos A = \dfrac{4}{5}\): using Pythagoras, \(\sin A = \dfrac{3}{5}\)
Since \(\cos B = \dfrac{3}{5}\): \(\sin B = \dfrac{4}{5}\)
Step 2 — Find sin C:
\(\sin C = \sin(A+B) = \sin A\cos B + \cos A\sin B\quad[\because A+B+C=180°]\)
\(= \dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{9}{25} + \dfrac{16}{25} = \dfrac{25}{25} = 1\)
\(\therefore \sin C = 1 \Rightarrow C = 90°\)
Step 3 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{3}{5}} = \dfrac{b}{\frac{4}{5}} = \dfrac{c}{1}\)
\(\Rightarrow \dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{5}\)
Step 1 — Find sin A and sin B:
Since \(\cos A = \dfrac{4}{5}\): using Pythagoras, \(\sin A = \dfrac{3}{5}\)
Since \(\cos B = \dfrac{3}{5}\): \(\sin B = \dfrac{4}{5}\)
Step 2 — Find sin C:
\(\sin C = \sin(A+B) = \sin A\cos B + \cos A\sin B\quad[\because A+B+C=180°]\)
\(= \dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{9}{25} + \dfrac{16}{25} = \dfrac{25}{25} = 1\)
\(\therefore \sin C = 1 \Rightarrow C = 90°\)
Step 3 — Apply Sine Rule:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{3}{5}} = \dfrac{b}{\frac{4}{5}} = \dfrac{c}{1}\)
\(\Rightarrow \dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{5}\)
✅ \(a:b:c = 3:4:5\)
Q2(i)
Find \(A,B,C\) given \(a = 2,\ b = \sqrt{2},\ c = \sqrt{3}+1\)
Given: \(a = 2,\ b = \sqrt{2},\ c = \sqrt{3}+1\)
Step 1 — Find cos A (Cosine Law):
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{(\sqrt{2})^2+(\sqrt{3}+1)^2-(2)^2}{2 \times \sqrt{2} \times (\sqrt{3}+1)}\)
\(= \dfrac{2 + 3 + 2\sqrt{3} + 1 – 4}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{2 + 2\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{2(\sqrt{3}+1)}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{1}{\sqrt{2}}\)
\(\therefore \cos A = \cos 45° \Rightarrow A = 45°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{(\sqrt{3}+1)^2+4-2}{2(\sqrt{3}+1) \cdot 2} = \dfrac{3+2\sqrt{3}+1+2}{4(\sqrt{3}+1)} = \dfrac{6+2\sqrt{3}}{4(\sqrt{3}+1)}\)
\(= \dfrac{2\sqrt{3}(\sqrt{3}+1)}{4(\sqrt{3}+1)} = \dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{\sqrt{2}} \cdot \cdots = \cos 30°\)
\(\therefore B = 30°\)
Step 3 — Find C:
\(A + B + C = 180° \Rightarrow 45° + 30° + C = 180°\)
Step 1 — Find cos A (Cosine Law):
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{(\sqrt{2})^2+(\sqrt{3}+1)^2-(2)^2}{2 \times \sqrt{2} \times (\sqrt{3}+1)}\)
\(= \dfrac{2 + 3 + 2\sqrt{3} + 1 – 4}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{2 + 2\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{2(\sqrt{3}+1)}{2\sqrt{2}(\sqrt{3}+1)} = \dfrac{1}{\sqrt{2}}\)
\(\therefore \cos A = \cos 45° \Rightarrow A = 45°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{(\sqrt{3}+1)^2+4-2}{2(\sqrt{3}+1) \cdot 2} = \dfrac{3+2\sqrt{3}+1+2}{4(\sqrt{3}+1)} = \dfrac{6+2\sqrt{3}}{4(\sqrt{3}+1)}\)
\(= \dfrac{2\sqrt{3}(\sqrt{3}+1)}{4(\sqrt{3}+1)} = \dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{\sqrt{2}} \cdot \cdots = \cos 30°\)
\(\therefore B = 30°\)
Step 3 — Find C:
\(A + B + C = 180° \Rightarrow 45° + 30° + C = 180°\)
✅ \(A = 45°,\ B = 30°,\ C = 105°\)
Q2(ii)
Find \(A,B,C\) given \(a = 3+\sqrt{3},\ b = 2\sqrt{3},\ c = \sqrt{6}\)
Given: \(a = 3+\sqrt{3},\ b = 2\sqrt{3},\ c = \sqrt{6}\)
Step 1 — Find cos A:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{12+6-(3+\sqrt{3})^2}{2 \cdot 2\sqrt{3} \cdot \sqrt{6}}\)
\(= \dfrac{18-(9+6\sqrt{3}+3)}{4\sqrt{18}} = \dfrac{18-12-6\sqrt{3}}{12\sqrt{2}} = \dfrac{6(1-\sqrt{3})}{12\sqrt{2}} = \dfrac{1-\sqrt{3}}{2\sqrt{2}}\)
\(\therefore \cos A = \cos 105° \Rightarrow A = 105°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{6+(3+\sqrt{3})^2-12}{2\sqrt{6}(3+\sqrt{3})} = \dfrac{6+6\sqrt{3}}{2\sqrt{6}(3+\sqrt{3})}\)
\(= \dfrac{6(1+\sqrt{3})}{2\sqrt{6} \cdot \sqrt{3}(\sqrt{3}+1)} = \dfrac{6}{2\sqrt{18}} = \dfrac{6}{6\sqrt{2}} = \dfrac{1}{\sqrt{2}}\)
\(\therefore \cos B = \cos 45° \Rightarrow B = 45°\)
Step 3: \(C = 180° – (105°+45°) = 30°\)
Step 1 — Find cos A:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{12+6-(3+\sqrt{3})^2}{2 \cdot 2\sqrt{3} \cdot \sqrt{6}}\)
\(= \dfrac{18-(9+6\sqrt{3}+3)}{4\sqrt{18}} = \dfrac{18-12-6\sqrt{3}}{12\sqrt{2}} = \dfrac{6(1-\sqrt{3})}{12\sqrt{2}} = \dfrac{1-\sqrt{3}}{2\sqrt{2}}\)
\(\therefore \cos A = \cos 105° \Rightarrow A = 105°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{6+(3+\sqrt{3})^2-12}{2\sqrt{6}(3+\sqrt{3})} = \dfrac{6+6\sqrt{3}}{2\sqrt{6}(3+\sqrt{3})}\)
\(= \dfrac{6(1+\sqrt{3})}{2\sqrt{6} \cdot \sqrt{3}(\sqrt{3}+1)} = \dfrac{6}{2\sqrt{18}} = \dfrac{6}{6\sqrt{2}} = \dfrac{1}{\sqrt{2}}\)
\(\therefore \cos B = \cos 45° \Rightarrow B = 45°\)
Step 3: \(C = 180° – (105°+45°) = 30°\)
✅ \(A = 105°,\ B = 45°,\ C = 30°\)
Q2(iii)
Find \(A,B,C\) given \(a:b:c = 2:\sqrt{6}:(\sqrt{3}+1)\)
Given: In \(\triangle ABC\), \(a:b:c = 2:\sqrt{6}:(\sqrt{3}+1)\)
Let \(a = 2K,\ b = \sqrt{6}K,\ c = (\sqrt{3}+1)K\)
Step 1 — Find cos A:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{6K^2+(\sqrt{3}+1)^2K^2-4K^2}{2\sqrt{6}K(\sqrt{3}+1)K}\)
\(= \dfrac{6+(4+2\sqrt{3})-4}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{\sqrt{3}}{\sqrt{6}} = \dfrac{1}{\sqrt{2}}\)
\(\therefore A = 45°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{(4+2\sqrt{3})+4-6}{4(\sqrt{3}+1)} = \dfrac{2+2\sqrt{3}}{4(\sqrt{3}+1)} = \dfrac{2(1+\sqrt{3})}{4(1+\sqrt{3})} = \dfrac{1}{2}\)
\(\therefore B = 60°\)
Step 3: \(C = 180°-(45°+60°) = 75°\)
Let \(a = 2K,\ b = \sqrt{6}K,\ c = (\sqrt{3}+1)K\)
Step 1 — Find cos A:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{6K^2+(\sqrt{3}+1)^2K^2-4K^2}{2\sqrt{6}K(\sqrt{3}+1)K}\)
\(= \dfrac{6+(4+2\sqrt{3})-4}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \dfrac{\sqrt{3}}{\sqrt{6}} = \dfrac{1}{\sqrt{2}}\)
\(\therefore A = 45°\)
Step 2 — Find cos B:
\(\cos B = \dfrac{c^2+a^2-b^2}{2ca} = \dfrac{(4+2\sqrt{3})+4-6}{4(\sqrt{3}+1)} = \dfrac{2+2\sqrt{3}}{4(\sqrt{3}+1)} = \dfrac{2(1+\sqrt{3})}{4(1+\sqrt{3})} = \dfrac{1}{2}\)
\(\therefore B = 60°\)
Step 3: \(C = 180°-(45°+60°) = 75°\)
✅ \(A = 45°,\ B = 60°,\ C = 75°\)
Q3(i)
Find \(a,b,A\) given \(C = 30°,\ B = 45°,\ c = 6\sqrt{2}\)
Given: \(C = 30°,\ B = 45°,\ c = 6\sqrt{2}\)
Step 1 — Find A:
\(A = 180°-(B+C) = 180°-(45°+30°) = 105°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{b}{\frac{1}{\sqrt{2}}} = \dfrac{6\sqrt{2}}{\frac{1}{2}} = 12\sqrt{2}\)
Step 3 — Solve for a and b:
Taking 1st and 3rd ratios: \(a = 12\sqrt{2} \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = 6(\sqrt{3}+1)\)
Taking 2nd and 3rd ratios: \(b = 12\sqrt{2} \times \dfrac{1}{\sqrt{2}} = 12\)
Step 1 — Find A:
\(A = 180°-(B+C) = 180°-(45°+30°) = 105°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{a}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{b}{\frac{1}{\sqrt{2}}} = \dfrac{6\sqrt{2}}{\frac{1}{2}} = 12\sqrt{2}\)
Step 3 — Solve for a and b:
Taking 1st and 3rd ratios: \(a = 12\sqrt{2} \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = 6(\sqrt{3}+1)\)
Taking 2nd and 3rd ratios: \(b = 12\sqrt{2} \times \dfrac{1}{\sqrt{2}} = 12\)
✅ \(A = 105°,\ a = 6(\sqrt{3}+1),\ b = 12\)
Q3(ii)
Find \(b,C,c\) given \(A = 60°,\ B = 75°,\ a = 2\sqrt{3}\)
Given: \(A = 60°,\ B = 75°,\ a = 2\sqrt{3}\)
Step 1 — Find C:
\(C = 180°-(A+B) = 180°-(60°+75°) = 45°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = \dfrac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{c}{\frac{1}{\sqrt{2}}}\)
\(\Rightarrow 4 = \dfrac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{c}{\frac{1}{\sqrt{2}}}\)
Step 3: \(b = 4 \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \dfrac{2(\sqrt{3}+1)}{\sqrt{2}} = \sqrt{6}+\sqrt{2}\), \(c = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}\)
Step 1 — Find C:
\(C = 180°-(A+B) = 180°-(60°+75°) = 45°\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = \dfrac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{c}{\frac{1}{\sqrt{2}}}\)
\(\Rightarrow 4 = \dfrac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \dfrac{c}{\frac{1}{\sqrt{2}}}\)
Step 3: \(b = 4 \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \dfrac{2(\sqrt{3}+1)}{\sqrt{2}} = \sqrt{6}+\sqrt{2}\), \(c = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}\)
✅ \(C = 45°,\ b = \sqrt{6}+\sqrt{2},\ c = 2\sqrt{2}\)
Q3(iii)
Find \(a,c,C\) given \(A = 30°,\ B = 45°,\ b = 2\)
Given: \(A = 30°,\ B = 45°,\ b = 2\)
Step 1 — Find C:
\(C = 180°-(A+B) = 180°-(30°+45°) = 105°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin 30°} = \dfrac{b}{\sin 45°} = \dfrac{c}{\sin 105°}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = \dfrac{2}{\frac{1}{\sqrt{2}}} = \dfrac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = 2\sqrt{2}\)
Step 3: \(a = 2\sqrt{2} \times \dfrac{1}{2} = \sqrt{2}\), \(c = 2\sqrt{2} \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \sqrt{3}+1\)
Step 1 — Find C:
\(C = 180°-(A+B) = 180°-(30°+45°) = 105°\)
Step 2 — Apply Sine Rule:
\(\dfrac{a}{\sin 30°} = \dfrac{b}{\sin 45°} = \dfrac{c}{\sin 105°}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = \dfrac{2}{\frac{1}{\sqrt{2}}} = \dfrac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\)
\(\Rightarrow \dfrac{a}{\frac{1}{2}} = 2\sqrt{2}\)
Step 3: \(a = 2\sqrt{2} \times \dfrac{1}{2} = \sqrt{2}\), \(c = 2\sqrt{2} \times \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \sqrt{3}+1\)
✅ \(C = 105°,\ a = \sqrt{2},\ c = \sqrt{3}+1\)
Q4(i)
Find \(A\) and \(B\) given \(a = 2,\ b = 4,\ C = 60°\)
Given: \(a = 2,\ b = 4,\ C = 60°\)
Step 1 — Find c using Cosine Law:
\(\cos C = \dfrac{a^2+b^2-c^2}{2ab} \Rightarrow \dfrac{1}{2} = \dfrac{4+16-c^2}{2 \cdot 2 \cdot 4} = \dfrac{20-c^2}{16}\)
\(\Rightarrow 8 = 20 – c^2 \Rightarrow c^2 = 12 \therefore c = 2\sqrt{3}\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{2}{\sin A} = \dfrac{4}{\sin B} = \dfrac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4\)
\(\Rightarrow \sin A = \dfrac{1}{2} \Rightarrow A = 30°\)
\(\Rightarrow \sin B = 1 \Rightarrow B = 90°\)
Step 1 — Find c using Cosine Law:
\(\cos C = \dfrac{a^2+b^2-c^2}{2ab} \Rightarrow \dfrac{1}{2} = \dfrac{4+16-c^2}{2 \cdot 2 \cdot 4} = \dfrac{20-c^2}{16}\)
\(\Rightarrow 8 = 20 – c^2 \Rightarrow c^2 = 12 \therefore c = 2\sqrt{3}\)
Step 2 — Apply Sine Law:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
\(\Rightarrow \dfrac{2}{\sin A} = \dfrac{4}{\sin B} = \dfrac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4\)
\(\Rightarrow \sin A = \dfrac{1}{2} \Rightarrow A = 30°\)
\(\Rightarrow \sin B = 1 \Rightarrow B = 90°\)
✅ \(A = 30°,\ B = 90°,\ c = 2\sqrt{3}\)
Q4(ii)
Find \(a,B,C\) given \(A = 60°,\ b = \sqrt{3}-1,\ c = \sqrt{3}+1\)
Given: \(A = 60°,\ b = \sqrt{3}-1,\ c = \sqrt{3}+1\)
Step 1 — Find a using Cosine Law:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} \Rightarrow \dfrac{1}{2} = \dfrac{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2-a^2}{2(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(\Rightarrow \dfrac{1}{2} = \dfrac{(4-2\sqrt{3})+(4+2\sqrt{3})-a^2}{2(3-1)} = \dfrac{8-a^2}{4}\)
\(\Rightarrow 2 = 8-a^2 \Rightarrow a^2 = 6 \therefore a = \sqrt{6}\)
Step 2 — Apply Sine Rule to find B and C:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{\sqrt{6}}{\frac{\sqrt{3}}{2}} = \dfrac{\sqrt{3}-1}{\sin B}\)
\(\Rightarrow \sin B = \dfrac{(\sqrt{3}-1)\sqrt{3}}{2\sqrt{6}} = \dfrac{\sqrt{3}-1}{2\sqrt{2}} = \sin 15°\)
\(\therefore B = 15°\)
\(\sin C = \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \sin 75°\text{ or }\sin 105°\), take \(C = 105°\) (since \(75°\) not possible)
Step 1 — Find a using Cosine Law:
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} \Rightarrow \dfrac{1}{2} = \dfrac{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2-a^2}{2(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(\Rightarrow \dfrac{1}{2} = \dfrac{(4-2\sqrt{3})+(4+2\sqrt{3})-a^2}{2(3-1)} = \dfrac{8-a^2}{4}\)
\(\Rightarrow 2 = 8-a^2 \Rightarrow a^2 = 6 \therefore a = \sqrt{6}\)
Step 2 — Apply Sine Rule to find B and C:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{\sqrt{6}}{\frac{\sqrt{3}}{2}} = \dfrac{\sqrt{3}-1}{\sin B}\)
\(\Rightarrow \sin B = \dfrac{(\sqrt{3}-1)\sqrt{3}}{2\sqrt{6}} = \dfrac{\sqrt{3}-1}{2\sqrt{2}} = \sin 15°\)
\(\therefore B = 15°\)
\(\sin C = \dfrac{\sqrt{3}+1}{2\sqrt{2}} = \sin 75°\text{ or }\sin 105°\), take \(C = 105°\) (since \(75°\) not possible)
✅ \(a = \sqrt{6},\ B = 15°,\ C = 105°\)
Q5(i)
Ambiguous case: \(a = 3,\ b = 3\sqrt{3},\ A = 30°\)
⚠ Ambiguous Case
Given: \(a = 3,\ b = 3\sqrt{3},\ A = 30°\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{3}{\frac{1}{2}} = \dfrac{3\sqrt{3}}{\sin B} \Rightarrow \sin B = \dfrac{\sqrt{3}}{2}\)
\(\therefore B = 60°\text{ or }B = 120°\quad\text{(Ambiguous case)}\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{3}{\frac{1}{2}} = \dfrac{3\sqrt{3}}{\sin B} \Rightarrow \sin B = \dfrac{\sqrt{3}}{2}\)
\(\therefore B = 60°\text{ or }B = 120°\quad\text{(Ambiguous case)}\)
Case 1: When B = 60°
\(C = 180°-(30°+60°) = 90°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = 6\)
\(\therefore B = 60°,\ C = 90°,\ c = 6\)
\(C = 180°-(30°+60°) = 90°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = 6\)
\(\therefore B = 60°,\ C = 90°,\ c = 6\)
Case 2: When B = 120°
\(C = 180°-(30°+120°) = 30°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = 3\)
\(\therefore B = 120°,\ C = 30°,\ c = 3\)
\(C = 180°-(30°+120°) = 30°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = 3\)
\(\therefore B = 120°,\ C = 30°,\ c = 3\)
✅ Two solutions: (1) \(B=60°,C=90°,c=6\) | (2) \(B=120°,C=30°,c=3\)
Q5(ii)
Ambiguous case: \(a = 2,\ b = \sqrt{3}+1,\ A = 45°\)
⚠ Ambiguous Case
Given: \(a = 2,\ b = \sqrt{3}+1,\ A = 45°\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{2}{\frac{1}{\sqrt{2}}} = \dfrac{\sqrt{3}+1}{\sin B}\)
\(\Rightarrow \sin B = \dfrac{\sqrt{3}+1}{2\sqrt{2}}\)
\(\therefore B = 75°\text{ or }B = 105°\quad\text{(Ambiguous case)}\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{2}{\frac{1}{\sqrt{2}}} = \dfrac{\sqrt{3}+1}{\sin B}\)
\(\Rightarrow \sin B = \dfrac{\sqrt{3}+1}{2\sqrt{2}}\)
\(\therefore B = 75°\text{ or }B = 105°\quad\text{(Ambiguous case)}\)
Case 1: When B = 75°
\(C = 180°-(45°+75°) = 60°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = \sqrt{6}\)
\(\therefore B = 75°,\ C = 60°,\ c = \sqrt{6}\)
\(C = 180°-(45°+75°) = 60°\)
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow c = \sqrt{6}\)
\(\therefore B = 75°,\ C = 60°,\ c = \sqrt{6}\)
Case 2: When B = 105°
\(C = 180°-(45°+105°) = 30°\)
\(c = \sqrt{2}\)
\(\therefore B = 105°,\ C = 30°,\ c = \sqrt{2}\)
\(C = 180°-(45°+105°) = 30°\)
\(c = \sqrt{2}\)
\(\therefore B = 105°,\ C = 30°,\ c = \sqrt{2}\)
✅ Two solutions: (1) \(B=75°,C=60°,c=\sqrt{6}\) | (2) \(B=105°,C=30°,c=\sqrt{2}\)
Q5(iii)
Find \(B,C,c\) given \(A = 30°,\ a = 6,\ b = 4\)
Given: \(A = 30°,\ a = 6,\ b = 4\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{6}{\frac{1}{2}} = \dfrac{4}{\sin B} \Rightarrow \sin B = \dfrac{1}{3}\)
\(\therefore B \approx 19.47°\quad\text{(No two solutions — unique triangle)}\)
Step 2 — Find C:
\(C = 180°-(30°+19.47°) \approx 130.53°\)
Step 3 — Find c:
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow \dfrac{6}{\frac{1}{2}} = \dfrac{c}{\sin(130.53°)} \Rightarrow c = 9\)
Step 1 — Find sin B:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{6}{\frac{1}{2}} = \dfrac{4}{\sin B} \Rightarrow \sin B = \dfrac{1}{3}\)
\(\therefore B \approx 19.47°\quad\text{(No two solutions — unique triangle)}\)
Step 2 — Find C:
\(C = 180°-(30°+19.47°) \approx 130.53°\)
Step 3 — Find c:
\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow \dfrac{6}{\frac{1}{2}} = \dfrac{c}{\sin(130.53°)} \Rightarrow c = 9\)
✅ \(B \approx 19.47°,\ C \approx 130.53°,\ c = 9\)
📌 Summary of Key Results — Exercise 7.1:
• Q1(i): \(a:b:c = \sqrt{3}+1:\sqrt{3}-1:\sqrt{6}\)
• Q1(ii): \(a:c = 2:\sqrt{3}+1\)
• Q1(iii): \(a:b:c = \sqrt{3}:1:2\), and \(A=60°, B=30°, C=90°\)
• Q1(v): \(a:b:c = 3:4:5\)
• Q2(i): \(A=45°, B=30°, C=105°\)
• Q2(iii): \(A=45°, B=60°, C=75°\)
• Q4(i): \(A=30°, B=90°, c=2\sqrt{3}\)
• Q5(i) Ambiguous: Two solutions — \((B=60°,C=90°,c=6)\) and \((B=120°,C=30°,c=3)\)
• Q1(i): \(a:b:c = \sqrt{3}+1:\sqrt{3}-1:\sqrt{6}\)
• Q1(ii): \(a:c = 2:\sqrt{3}+1\)
• Q1(iii): \(a:b:c = \sqrt{3}:1:2\), and \(A=60°, B=30°, C=90°\)
• Q1(v): \(a:b:c = 3:4:5\)
• Q2(i): \(A=45°, B=30°, C=105°\)
• Q2(iii): \(A=45°, B=60°, C=75°\)
• Q4(i): \(A=30°, B=90°, c=2\sqrt{3}\)
• Q5(i) Ambiguous: Two solutions — \((B=60°,C=90°,c=6)\) and \((B=120°,C=30°,c=3)\)
📄 Solution of Triangle Class 12 Mathematics Notes PDF
This PDF provides the solutions of every question from Class 12 Solution of Triangle. This chapter has only one exercise. Notes are updated according to the latest syllabus of 2080. Please do not share this PDF on any website or social platform without permission.
This PDF provides the solutions of every question from Class 12 Solution of Triangle. This chapter has only one exercise. Notes are updated according to the latest syllabus of 2080. Please do not share this PDF on any website or social platform without permission.