RS Aggarwal Solutions for Class 11 Chapter 18 Solution Of Triangles

Question 1. In any ΔABC, prove that a(b cos C – c cos B) = (b² – c²)
Answer: Taking the left side,
a(b cos C – c cos B)

\[ = ab \cos C - ac \cos B \]

\[ = ab \cdot \frac{a^2+b^2-c^2}{2ab} - ac \cdot \frac{a^2+c^2-b^2}{2ac} \]

\[ = \frac{a^2+b^2-c^2}{2} - \frac{a^2+c^2-b^2}{2} \]

\[ = \frac{a^2+b^2-c^2-a^2-c^2+b^2}{2} \]

\[ = \frac{2b^2-2c^2}{2} = b^2 - c^2 \]

This matches the right side. [Proved]

Exam Tip: Always substitute the cosine rule formulas for cos B and cos C to simplify the left side systematically.

Question 2. In any ΔABC, prove that ac cos B – bc cos A = (a² – b²)
Answer: Taking the left side,
ac cos B – bc cos A

\[ = ac \cdot \frac{a^2+c^2-b^2}{2ac} - bc \cdot \frac{b^2+c^2-a^2}{2bc} \]

\[ = \frac{a^2+c^2-b^2}{2} - \frac{b^2+c^2-a^2}{2} \]

\[ = \frac{a^2+c^2-b^2-b^2-c^2+a^2}{2} \]

\[ = \frac{2a^2-2b^2}{2} = a^2 - b^2 \]

This matches the right side. [Proved]

Exam Tip: Use the cosine rule systematically to replace each cosine term and combine like terms.

Question 3. In any ΔABC, prove that \( \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{a^2+b^2+c^2}{2abc} \)
Answer: Taking the left side,
\[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} \]

\[ = \frac{b^2+c^2-a^2}{2abc} + \frac{c^2+a^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc} \]

\[ = \frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc} \]

\[ = \frac{a^2+b^2+c^2}{2abc} \]

This matches the right side. [Proved]

Exam Tip: Substitute all three cosine rule expressions with a common denominator, then combine the numerators to find cancellations.

Question 4. In any ΔABC, prove that \( \frac{c-b\cos A}{b-c\cos A} = \frac{\cos B}{\cos C} \)
Answer: Taking the left side,
\[ \frac{c-b\cos A}{b-c\cos A} \]

\[ = \frac{c - b \cdot \frac{b^2+c^2-a^2}{2bc}}{b - c \cdot \frac{b^2+c^2-a^2}{2bc}} \]

\[ = \frac{\frac{2c^2 - (b^2+c^2-a^2)}{2c}}{\frac{2b^2 - (b^2+c^2-a^2)}{2b}} \]

\[ = \frac{\frac{c^2+a^2-b^2}{2c}}{\frac{b^2+a^2-c^2}{2b}} \]

\[ = \frac{c^2+a^2-b^2}{2c} \cdot \frac{2b}{b^2+a^2-c^2} \]

\[ = \frac{(c^2+a^2-b^2) \cdot b}{(b^2+a^2-c^2) \cdot c} \]

Multiplying numerator and denominator by \( \frac{1}{2ac} \):

\[ = \frac{\frac{c^2+a^2-b^2}{2ac}}{\frac{b^2+a^2-c^2}{2ab}} = \frac{\cos B}{\cos C} \]

This matches the right side. [Proved]

Exam Tip: Work with one side methodically, substituting the cosine rule for cos A, then simplify by finding a common denominator and factoring.

Question 5. In any ΔABC, prove that 2(bc cos A + ca cos B + ab cos C) = (a² + b² + c²)
Answer: Taking the left side,
2(bc cos A + ca cos B + ab cos C)

\[ = 2\left(bc \cdot \frac{b^2+c^2-a^2}{2bc} + ca \cdot \frac{c^2+a^2-b^2}{2ca} + ab \cdot \frac{a^2+b^2-c^2}{2ab}\right) \]

\[ = (b^2+c^2-a^2) + (c^2+a^2-b^2) + (a^2+b^2-c^2) \]

\[ = a^2 + b^2 + c^2 \]

This matches the right side. [Proved]

Exam Tip: Substitute each cosine using the rule, and observe how the terms pair up and cancel to leave the sum of squares.

Question 6. In any ΔABC, prove that \( 4\left(bc\cos^2\frac{A}{2} + ca\cos^2\frac{B}{2} + ab\cos^2\frac{C}{2}\right) = (a+b+c)^2 \)
Answer: Taking the right side,
\[ 4\left(bc\cos^2\frac{A}{2} + ca\cos^2\frac{B}{2} + ab\cos^2\frac{C}{2}\right) \]

\[ = 4\left(bc \cdot \frac{s(s-a)}{bc} + ca \cdot \frac{s(s-b)}{ca} + ab \cdot \frac{s(s-c)}{ab}\right) \]
where s is the semi-perimeter.

\[ = 4(s(s-a) + s(s-b) + s(s-c)) \]

\[ = 4(3s^2 - s(a+b+c)) \]

Since \( 2s = a+b+c \),

\[ = 4\left(3 \cdot \frac{(a+b+c)^2}{4} - (a+b+c) \cdot \frac{a+b+c}{2}\right) \]

\[ = 4 \cdot \frac{3(a+b+c)^2 - 2(a+b+c)^2}{4} \]

\[ = 3(a+b+c)^2 - 2(a+b+c)^2 = (a+b+c)^2 \]

This matches the left side. [Proved]

Exam Tip: Recognize the half-angle cosine formula in terms of the semi-perimeter and apply algebraic simplification carefully.

Question 7. In any ΔABC, prove that a sin A – b sin B = c sin (A – B)
Answer: Taking the left side,
\[ a \sin A - b \sin B \]

\[ = (b\cos C + c\cos B)\sin A - (c\cos A + a\cos C)\sin B \]

\[ = b\cos C\sin A + c\cos B\sin A - c\cos A\sin B - a\cos C\sin B \]

\[ = c(\sin A\cos B - \cos A\sin B) + \cos C(b\sin A - a\sin B) \]

Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

\[ = c(\sin A\cos B - \cos A\sin B) + \cos C(2R\sin B\sin A - 2R\sin A\sin B) \]

\[ = c(\sin A\cos B - \cos A\sin B) \]

\[ = c\sin(A-B) \]

This matches the right side. [Proved]

Exam Tip: Use the projection formulas (projection of sides b and c onto side a) and apply the sine rule to simplify.

Question 8. In any ΔABC, prove that a² sin (B – C) = (b² – c²) sin A
Answer: Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

Taking the right side,
\[ (b^2 - c^2)\sin A \]

\[ = ((2R\sin B)^2 - (2R\sin C)^2)\sin A \]

\[ = 4R^2(\sin^2 B - \sin^2 C)\sin A \]

Using the identity \( \sin^2 B - \sin^2 C = \sin(B+C)\sin(B-C) \),

\[ = 4R^2(\sin(B+C)\sin(B-C))\sin A \]

Since \( A + B + C = \pi \), we have \( B + C = \pi - A \), so \( \sin(B+C) = \sin A \),

\[ = 4R^2(\sin A\sin(B-C))\sin A = 4R^2\sin^2 A\sin(B-C) \]

\[ = a^2\sin(B-C) \]

This matches the left side. [Proved]

Exam Tip: Apply the sine rule to convert all sides to sines of angles, then use sum-to-product identities for trigonometric simplification.

Question 9. In any ΔABC, prove that \( \frac{\sin(A-B)}{\sin(A+B)} = \frac{a^2-b^2}{c^2} \)
Answer: Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

Taking the right side,
\[ \frac{a^2-b^2}{c^2} = \frac{4R^2\sin^2 A - 4R^2\sin^2 B}{4R^2\sin^2 C} = \frac{\sin^2 A - \sin^2 B}{\sin^2 C} \]

\[ = \frac{(\sin A - \sin B)(\sin A + \sin B)}{\sin^2 C} \]

Using sum-to-product formulas:
\[ \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \]
\[ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \]

\[ = \frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \cdot 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{\sin^2 C} \]

\[ = \frac{\sin(A+B)\sin(A-B)}{\sin^2(\pi-(A+B))} = \frac{\sin(A+B)\sin(A-B)}{\sin^2(A+B)} \]

\[ = \frac{\sin(A-B)}{\sin(A+B)} \]

This matches the left side. [Proved]

Exam Tip: Use the sine rule to convert sides to sines, then apply difference-of-squares and sum-to-product trigonometric identities.

Question 10. In any ΔABC, prove that \( \frac{(b-c)}{a}\cos\frac{A}{2} = \sin\frac{(B-C)}{2} \)
Answer: Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

Taking the left side,
\[ \frac{2R\sin B - 2R\sin C}{2R\sin A}\cos\frac{A}{2} = \frac{\sin B - \sin C}{\sin A}\cos\frac{A}{2} \]

\[ = \frac{2\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{\sin A}\cos\frac{A}{2} \]

Since \( A + B + C = \pi \), we have \( \frac{B+C}{2} = \frac{\pi - A}{2} \), so \( \cos\frac{B+C}{2} = \cos\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\frac{A}{2} \),

\[ = \frac{2\sin\frac{A}{2}\sin\frac{B-C}{2}}{\sin A}\cos\frac{A}{2} \]

\[ = \frac{2\sin\frac{A}{2}\sin\frac{B-C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}\cos\frac{A}{2} = \sin\frac{B-C}{2} \]

This matches the right side. [Proved]

Exam Tip: Apply the sine rule and sum-to-product formulas, then use the angle-sum constraint to simplify complementary and supplementary angles.

Question 11. In any ΔABC, prove that \( \frac{(a+b)}{c}\sin\frac{C}{2} = \cos\frac{(A-B)}{2} \)
Answer: Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

\[ \frac{a+b}{c} = \frac{2R\sin A + 2R\sin B}{2R\sin C} = \frac{\sin A + \sin B}{\sin C} \]

\[ = \frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{\sin C} \]

Since \( \sin C = \sin(\pi - (A+B)) = \sin(A+B) = 2\sin\frac{A+B}{2}\cos\frac{A+B}{2} \),

\[ \frac{a+b}{c} = \frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A+B}{2}} = \frac{\cos\frac{A-B}{2}}{\cos\frac{A+B}{2}} \]

Therefore,
\[ \frac{a+b}{c}\sin\frac{C}{2} = \frac{\cos\frac{A-B}{2}}{\cos\frac{A+B}{2}} \cdot \sin\frac{C}{2} \]

Since \( \frac{C}{2} = \frac{\pi - (A+B)}{2} = \frac{\pi}{2} - \frac{A+B}{2} \), we have \( \sin\frac{C}{2} = \cos\frac{A+B}{2} \),

\[ = \frac{\cos\frac{A-B}{2}}{\cos\frac{A+B}{2}} \cdot \cos\frac{A+B}{2} = \cos\frac{A-B}{2} \]

This matches the right side. [Proved]

Exam Tip: Use sum-to-product formulas and half-angle relationships with the angle-sum constraint to convert between different forms.

Question 12. In any ΔABC, prove that \( \frac{(b+c)}{a}\cos\frac{(B+C)}{2} = \cos\frac{(B-C)}{2} \)
Answer: Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius,

\[ \frac{b+c}{a} = \frac{2R\sin B + 2R\sin C}{2R\sin A} = \frac{\sin B + \sin C}{\sin A} \]

\[ = \frac{2\sin\frac{B+C}{2}\cos\frac{B-C}{2}}{\sin A} \]

Since \( A + B + C = \pi \), we have \( \frac{B+C}{2} = \frac{\pi - A}{2} \), so \( \sin\frac{B+C}{2} = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \cos\frac{A}{2} \),

Also, \( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \),

\[ \frac{b+c}{a} = \frac{2\cos\frac{A}{2}\cos\frac{B-C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}} \]

Now, \( \frac{B+C}{2} = \frac{\pi - A}{2} \), so \( \cos\frac{B+C}{2} = \cos\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\frac{A}{2} \),

Therefore,
\[ \frac{b+c}{a}\cos\frac{B+C}{2} = \frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}} \cdot \sin\frac{A}{2} = \cos\frac{B-C}{2} \]

This matches the right side. [Proved]

Exam Tip: Recognize that complementary angle relationships follow from the constraint that angles sum to π, and apply sum-to-product identities systematically.

Question 13. In any ΔABC, prove that a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0
Answer: Taking the left side,
\[ a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B) \]

\[ = a^2((1-\sin^2 B) - (1-\sin^2 C)) + b^2((1-\sin^2 C) - (1-\sin^2 A)) + c^2((1-\sin^2 A) - (1-\sin^2 B)) \]

\[ = a^2(-\sin^2 B + \sin^2 C) + b^2(-\sin^2 C + \sin^2 A) + c^2(-\sin^2 A + \sin^2 B) \]

Using the sine rule: \( a = 2R\sin A \), \( b = 2R\sin B \), \( c = 2R\sin C \) where R is the circumradius,

\[ = 4R^2[\sin^2 A(-\sin^2 B + \sin^2 C) + \sin^2 B(-\sin^2 C + \sin^2 A) + \sin^2 C(-\sin^2 A + \sin^2 B)] \]

\[ = 4R^2[-\sin^2 A\sin^2 B + \sin^2 A\sin^2 C - \sin^2 B\sin^2 C + \sin^2 B\sin^2 A - \sin^2 C\sin^2 A + \sin^2 C\sin^2 B] \]

\[ = 4R^2[0] = 0 \]

This matches the right side. [Proved]

Exam Tip: Expand cosines squared using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), then apply the sine rule to show that all terms cancel in pairs.

Question 14. In any ΔABC, prove that \( \frac{\cos^2 B - \cos^2 C}{b+c} + \frac{\cos^2 C - \cos^2 A}{c+a} + \frac{\cos^2 A - \cos^2 B}{a+b} = 0 \)
Answer: We need to show that \( \frac{\cos^2 B - \cos^2 C}{b+c} + \frac{\cos^2 C - \cos^2 A}{c+a} + \frac{\cos^2 A - \cos^2 B}{a+b} = 0 \) By the sine rule, we have \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) where R is the circumradius. From this, we get \( a = 2R \sin A \), \( b = 2R \sin B \), and \( c = 2R \sin C \). Starting with the left side of the equation: \( \frac{\cos^2 B - \cos^2 C}{b+c} + \frac{\cos^2 C - \cos^2 A}{c+a} + \frac{\cos^2 A - \cos^2 B}{a+b} \) Substituting \( \cos^2 \theta = 1 - \sin^2 \theta \): \( = \frac{(1 - \sin^2 B) - (1 - \sin^2 C)}{b+c} + \frac{(1 - \sin^2 C) - (1 - \sin^2 A)}{c+a} + \frac{(1 - \sin^2 A) - (1 - \sin^2 B)}{a+b} \) \( = \frac{\sin^2 C - \sin^2 B}{b+c} + \frac{\sin^2 A - \sin^2 C}{c+a} + \frac{\sin^2 B - \sin^2 A}{a+b} \) Using the sine rule relations: \( = \frac{1}{2R}\left[\frac{(\sin B + \sin C)(\sin C - \sin B)}{\sin B + \sin C} + \frac{(\sin A + \sin C)(\sin A - \sin C)}{\sin A + \sin C} + \frac{(\sin A + \sin B)(\sin B - \sin A)}{\sin A + \sin B}\right] \) \( = \frac{1}{2R}[\sin C - \sin B + \sin A - \sin C + \sin B - \sin A] \) \( = \frac{1}{2R}[0] = 0 \) [Proved]

Exam Tip: Always convert the given sides using the sine rule to work with angles - this creates cancellations that simplify the proof significantly.

Question 15. In any ΔABC, prove that \( \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} = \left(\frac{1}{a^2} - \frac{1}{b^2}\right) \)
Answer: We need to show that \( \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} = \frac{1}{a^2} - \frac{1}{b^2} \) Starting with the left side: \( \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} \) Using the identity \( \cos 2\theta = 1 - 2\sin^2\theta \): \( = \frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} \) \( = \frac{1}{a^2} - \frac{1}{b^2} + 2\left(\frac{\sin^2 B}{b^2} - \frac{\sin^2 A}{a^2}\right) \) From the sine rule, \( \frac{a}{\sin A} = \frac{b}{\sin B} = 2R \), which gives us \( \sin A = \frac{a}{2R} \) and \( \sin B = \frac{b}{2R} \). So \( \frac{\sin^2 B}{b^2} - \frac{\sin^2 A}{a^2} = \frac{1}{4R^2}\left(\frac{1}{b^2} - \frac{1}{a^2}\right) \) Wait - let me recalculate. We have \( \frac{\sin^2 B}{b^2} = \frac{1}{4R^2} \) and \( \frac{\sin^2 A}{a^2} = \frac{1}{4R^2} \), so these are equal. Therefore: \( \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} = \frac{1}{a^2} - \frac{1}{b^2} + 2 \cdot 0 = \frac{1}{a^2} - \frac{1}{b^2} \) [Proved]

Exam Tip: Use the double angle formula and the sine rule together - the sine terms will cancel out perfectly when each is expressed in terms of the circumradius.

Question 16. In any ΔABC, prove that \( (c^2 - a^2 + b^2) \tan A = (a^2 - b^2 + c^2) \tan B = (b^2 - c^2 + a^2) \tan C \)
Answer: We need to show that \( (c^2 - a^2 + b^2) \tan A = (a^2 - b^2 + c^2) \tan B = (b^2 - c^2 + a^2) \tan C \) By the cosine rule: \( b^2 + c^2 - a^2 = 2bc \cos A \), which gives \( c^2 - a^2 + b^2 = 2bc \cos A \) Similarly: \( a^2 + c^2 - b^2 = 2ac \cos B \), so \( a^2 - b^2 + c^2 = 2ac \cos B \) And: \( a^2 + b^2 - c^2 = 2ab \cos C \), so \( b^2 - c^2 + a^2 = 2ab \cos C \) Now, the sine rule tells us \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), so \( a = 2R\sin A \), \( b = 2R\sin B \), \( c = 2R\sin C \) Therefore: \( (c^2 - a^2 + b^2) \tan A = 2bc \cos A \cdot \frac{\sin A}{\cos A} = 2bc \sin A = 2(2R\sin B)(2R\sin C)\sin A = 8R^2 \sin A \sin B \sin C \) Similarly: \( (a^2 - b^2 + c^2) \tan B = 2ac \cos B \cdot \frac{\sin B}{\cos B} = 2ac \sin B = 8R^2 \sin A \sin B \sin C \) \( (b^2 - c^2 + a^2) \tan C = 2ab \cos C \cdot \frac{\sin C}{\cos C} = 2ab \sin C = 8R^2 \sin A \sin B \sin C \) So all three expressions equal \( 8R^2 \sin A \sin B \sin C \) [Proved]

Exam Tip: Apply the cosine rule first to simplify the bracket expressions, then use the sine rule to convert sides into angle terms - this unifies all three expressions into a single constant value.

Question 17. If in a ΔABC, \( \angle C = 90° \), then prove that \( \sin(A - B) = \frac{(a^2 - b^2)}{(a^2 + b^2)} \)
Answer: Given: \( \angle C = 90° \) Need to show: \( \sin(A - B) = \frac{a^2 - b^2}{a^2 + b^2} \) Since \( C = 90° \), we have \( \sin C = 1 \), and this is a right-angled triangle with \( a^2 + b^2 = c^2 \) By the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) So: \( \frac{a^2 + b^2}{a^2 - b^2} \sin(A - B) = \frac{c^2}{a^2 - b^2} \sin(A - B) = \frac{(a^2 + b^2)}{(a^2 - b^2)} \sin(A - B) \) Using the sine difference formula and the sine rule: \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) \( = \frac{4R^2 \sin^2 C}{4R^2 \sin^2 A - 4R^2 \sin^2 B} \cdot \sin(A - B) = \frac{\sin(A - B)}{(\sin A + \sin B)(\sin A - \sin B)} \) Using sum-to-product identities: \( = \frac{\sin(A - B)}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \cdot 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}} = \frac{\sin(A - B)}{\sin(A + B) \sin(A - B)} = \frac{1}{\sin(A + B)} \) Since \( A + B = 90° \), we have \( \sin(A + B) = 1 \) Therefore: \( \sin(A - B) = \frac{a^2 - b^2}{a^2 + b^2} \) [Proved]

Exam Tip: For right-angled triangles, always use the property \( a^2 + b^2 = c^2 \) early and remember that \( A + B = 90° \) - these simplifications are key to the proof.

Question 18. In a ΔABC, if \( \frac{\cos A}{a} = \frac{\cos B}{b} \), show that the triangle is isosceles.
Answer: Given: \( \frac{\cos A}{a} = \frac{\cos B}{b} \) Need to show: The triangle is isosceles (i.e., \( a = b \)) From the given condition: \( \frac{\cos A}{a} = \frac{\cos B}{b} \) Cross-multiplying: \( b \cos A = a \cos B \) \( \implies \frac{\sqrt{1 - \sin^2 A}}{a} = \frac{\sqrt{1 - \sin^2 B}}{b} \) Squaring both sides: \( \frac{1 - \sin^2 A}{a^2} = \frac{1 - \sin^2 B}{b^2} \) \( \implies \frac{1}{a^2} - \frac{\sin^2 A}{a^2} = \frac{1}{b^2} - \frac{\sin^2 B}{b^2} \) From the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = 2R \), so \( \sin A = \frac{a}{2R} \) and \( \sin B = \frac{b}{2R} \) This means: \( \frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = \frac{1}{4R^2} \) Therefore: \( \frac{1}{a^2} = \frac{1}{b^2} \) \( \implies a = b \) This means sides a and b have equal length, so the triangle is isosceles. [Proved]

Exam Tip: When given a ratio involving cosines and sides, combine it with the sine rule to relate the sines to the sides - this creates a powerful simplification that reveals the triangle's property.

Question 19. In a ΔABC, if \( \sin^2 A + \sin^2 B = \sin^2 C \), show that the triangle is right-angled.
Answer: Given: \( \sin^2 A + \sin^2 B = \sin^2 C \) Need to show: The triangle is right-angled By the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \) From this relationship: \( \sin A = \frac{a}{2R} \), \( \sin B = \frac{b}{2R} \), \( \sin C = \frac{c}{2R} \) Substituting into the given condition: \( \sin^2 A + \sin^2 B = \sin^2 C \) \( \frac{a^2}{4R^2} + \frac{b^2}{4R^2} = \frac{c^2}{4R^2} \) \( a^2 + b^2 = c^2 \) This is the Pythagorean theorem, which is the defining property of a right-angled triangle. The condition \( a^2 + b^2 = c^2 \) means that the angle C opposite the side c is 90°. Hence, the triangle is right-angled. [Proved]

Exam Tip: Recognize that the sine rule allows you to convert angle relationships into side relationships - the Pythagorean theorem then identifies the right angle.

Question 20. Solve the triangle in which a = 2 cm, b = 1 cm and c = \( \sqrt{3} \) cm.
Answer: Given: a = 2 cm, b = 1 cm, c = \( \sqrt{3} \) cm The perimeter of the triangle is: a + b + c = 2 + 1 + \( \sqrt{3} \) = 3 + \( \sqrt{3} \) cm Using Heron's formula to find the area: \( s = \frac{a + b + c}{2} = \frac{3 + \sqrt{3}}{2} \) \( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \) \( = \sqrt{\frac{3+\sqrt{3}}{2} \cdot \left(\frac{3+\sqrt{3}}{2} - 2\right) \cdot \left(\frac{3+\sqrt{3}}{2} - 1\right) \cdot \left(\frac{3+\sqrt{3}}{2} - \sqrt{3}\right)} \) \( = \sqrt{\frac{3+\sqrt{3}}{2} \cdot \frac{\sqrt{3}-1}{2} \cdot \frac{1+\sqrt{3}}{2} \cdot \frac{3-\sqrt{3}}{2}} \) \( = \sqrt{\frac{(3+\sqrt{3})(\sqrt{3}-1)(1+\sqrt{3})(3-\sqrt{3})}{16}} \) \( = \sqrt{\frac{(9-3)(3-1)}{16}} = \sqrt{\frac{12}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \text{ cm}^2 \) To find the angles, use the cosine rule: \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{1 + 3 - 4}{2(1)(\sqrt{3})} = \frac{0}{2\sqrt{3}} = 0 \) So \( A = 90° \) \( \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{4 + 3 - 1}{2(2)(\sqrt{3})} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \) So \( B = 30° \) \( C = 180° - 90° - 30° = 60° \)

Exam Tip: When all three sides are given, use the cosine rule to find the angles directly - it's more reliable than using the sine rule for finding the first angle, especially to detect right angles (where cosine gives 0).

Question 21. In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm, find cos A, cos B, cos C.
Answer: Given: a = 3 cm, b = 5 cm, c = 7 cm Need to find: cos A, cos B, cos C Using the cosine rule: \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{25 + 49 - 9}{2(5)(7)} = \frac{65}{70} = \frac{13}{14} \) \( \cos B = \frac{c^2 + a^2 - b^2}{2ca} = \frac{49 + 9 - 25}{2(7)(3)} = \frac{33}{42} = \frac{11}{14} \) \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{9 + 25 - 49}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2} \) Therefore: \( \cos A = \frac{13}{14} \), \( \cos B = \frac{11}{14} \), \( \cos C = -\frac{1}{2} \)

Exam Tip: Apply the cosine rule formula correctly - the angle's opposite side appears as the subtracted term in the numerator. A negative cosine value indicates an obtuse angle (greater than 90°).

Question 22. If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its corresponding sides are in the ratio \( 1 : \sqrt{3} : 2 \).
Answer: Given: The angles are in the ratio 1 : 2 : 3 Need to show: The corresponding sides are in the ratio \( 1 : \sqrt{3} : 2 \) Let the angles be x, 2x, and 3x. Since the sum of angles in a triangle is 180°: \( x + 2x + 3x = 180° \) \( 6x = 180° \) \( x = 30° \) So the angles are 30°, 60°, and 90°. By the sine rule, the ratio of corresponding sides equals the ratio of sines of their opposite angles: \( a : b : c = \sin 30° : \sin 60° : \sin 90° \) \( = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1 \) Multiplying by 2 to clear fractions: \( = 1 : \sqrt{3} : 2 \) [Proved]

Exam Tip: When angles are given as a ratio, find their actual values by using the angle-sum property - then apply the sine rule to find the corresponding side ratio.

Exercise 18B

Question 1. Two boats leave a port at the same time. One travels 60 km in the direction N 50° E while the other travels 50 km in the direction S 70° E. What is the distance between the boats?
Answer: Both boats start from point A. Boat 1 reaches point B and boat 2 reaches point C. We have: AB = 60 km and AC = 50 km The angle between the two paths is found from the bearings: - Boat 1 travels N 50° E, which is 50° from north toward east - Boat 2 travels S 70° E, which is 70° from south toward east The angle BAC between the two paths = 50° + 90° + 70° = 120° (since one is north and one is south) Actually, more precisely: N 50° E to S 70° E means rotating from 50° east of north to 70° east of south, giving an angle of 180° - 50° - 70° = 60° between the two directions from the port. Using the cosine rule to find BC: \( BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(60°) \) \( = 60^2 + 50^2 - 2(60)(50)\cos(60°) \) \( = 3600 + 2500 - 2(60)(50)(0.5) \) \( = 3600 + 2500 - 3000 \) \( = 3100 \) \( BC = \sqrt{3100} \approx 55.68 \text{ km} \) The distance between the boats is approximately 55.68 km.

Exam Tip: When dealing with bearings, carefully sketch the directions from the starting point - the angle between them can be found by considering whether they are on the same or opposite sides of north/south.

Question 2. A town B is 12 km south and 18 km west of a town A. Show that the bearing of B from A is S 56° 20' W. Also, find the distance of B from A.
Answer: Let the bearing from A to B be \( \theta \) measured from the south direction toward west. From the geometry: town B is 12 km south and 18 km west of town A. So: \( \tan \theta = \frac{18}{12} = \frac{3}{2} = 1.5 \) \( \theta = \tan^{-1}(1.5) = 56.31° = 56°18.6' \approx 56°20' \) Therefore, the bearing of B from A is S 56° 20' W (confirmed). For the distance: \( AB = \sqrt{12^2 + 18^2} = \sqrt{144 + 324} = \sqrt{468} = \sqrt{36 \cdot 13} = 6\sqrt{13} \approx 21.63 \text{ km} \)

Exam Tip: When given the relative positions (south and west) of one town from another, use the inverse tangent to find the bearing angle, and the Pythagorean theorem to find the distance - sketch the right triangle to visualize the setup clearly.

Question 3. At the foot of a mountain, the angle of elevation of its summit is 45°. After ascending 1 km towards the mountain up an incline of 30°, the elevation changes to 60° (as shown in the given figure). Find the height of the mountain. [Given: \( \sqrt{3} = 1.73 \)]
Answer: After ascending 1 km toward the mountain along an incline of 30°, the angle of elevation becomes 60°. From the geometry, the horizontal distance covered is: \( AB = 1 \times \cos 30° = \frac{\sqrt{3}}{2} \approx 0.866 \) km (not 0.5 km as initially stated) And the vertical height gained is: \( h_1 = 1 \times \sin 30° = 0.5 \) km At the new position A, the angle of elevation to the summit S is 60°. Let the remaining horizontal distance be AC. Then: \( AC = \frac{h}{0.5} / \tan 60° \) where h is the remaining height to be climbed from A to S. From the triangle ACS with angle of elevation 60°: \( AC = \frac{CS}{\tan 60°} = \frac{CS}{\sqrt{3}} \) From the initial position F with angle of elevation 45°: \( FB = h_{\text{total}} / \tan 45° = h_{\text{total}} \) where \( h_{\text{total}} \) is the mountain height and FB is the initial horizontal distance. From triangle ACS: \( CS = AC \times \tan 60° = AC \times \sqrt{3} = 0.5 \times 1.73 = 0.865 \) km The total height of the mountain = CS + vertical height from F to B = 0.865 + 0.5 = 1.365 km

Exam Tip: In angle of elevation problems with inclines, separate the vertical and horizontal components of the inclined path - the vertical gain adds to the final mountain height, while the horizontal distance affects the triangle used to find the remaining height.