RBSE Solutions Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Exercise 14.1

 

Question 1. If in a triangle ABC, \( \angle A = 90^\circ \), a = 25 cm, b = 7 cm, then find all the trigonometrical ratios of \( \angle B \) and \( \angle C \).
Answer: We are given a right-angled triangle ABC where \( \angle A = 90^\circ \). The length of side 'a' (opposite to angle A, so BC) is 25 cm. The length of side 'b' (opposite to angle B, so AC) is 7 cm. We need to find the third side AB using the Pythagorean theorem.
We know that in a right-angled triangle, \( \text{hypotenuse}^2 = \text{perpendicular}^2 + \text{base}^2 \).
So, \( AB^2 + AC^2 = BC^2 \)
\( AB^2 + 7^2 = 25^2 \)
\( AB^2 + 49 = 625 \)
\( AB^2 = 625 - 49 \)
\( AB^2 = 576 \)
\( AB = \sqrt{576} \)
\( AB = 24 \text{ cm} \)
Now we have all three sides: \( AB = 24 \text{ cm} \), \( AC = 7 \text{ cm} \), and \( BC = 25 \text{ cm} \). We can find the trigonometric ratios for angles B and C.

For acute angle B:
Opposite side to B is AC = 7 cm.
Adjacent side to B is AB = 24 cm.
Hypotenuse is BC = 25 cm.
\( \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{7}{25} \)
\( \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{BC} = \frac{24}{25} \)
\( \tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{AB} = \frac{7}{24} \)
\( \csc B = \frac{1}{\sin B} = \frac{25}{7} \)
\( \sec B = \frac{1}{\cos B} = \frac{25}{24} \)
\( \cot B = \frac{1}{\tan B} = \frac{24}{7} \)

For acute angle C:
Opposite side to C is AB = 24 cm.
Adjacent side to C is AC = 7 cm.
Hypotenuse is BC = 25 cm.
\( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{BC} = \frac{24}{25} \)
\( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{7}{25} \)
\( \tan C = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{AC} = \frac{24}{7} \)
\( \csc C = \frac{1}{\sin C} = \frac{25}{24} \)
\( \sec C = \frac{1}{\cos C} = \frac{25}{7} \)
\( \cot C = \frac{1}{\tan C} = \frac{7}{24} \)
In simple words: First, we used Pythagoras' theorem to find the missing side of the triangle. Then, we used the lengths of all three sides to calculate the six trigonometry values (sin, cos, tan, csc, sec, cot) for both angle B and angle C. Remember, the 'opposite' and 'adjacent' sides change depending on which angle you are looking at.

🎯 Exam Tip: Always draw and label the right-angled triangle clearly. This helps correctly identify the opposite, adjacent, and hypotenuse sides for each angle, which is key for accurate ratio calculations.

 

Question 2. If in \( \triangle ABC \), \( \angle B = 90^\circ \), a = 12 cm, b = 13 cm, then find the trigonometrical ratios of \( \angle A \) and \( \angle C \).
Answer: We have a right-angled triangle ABC where \( \angle B = 90^\circ \). The length of side 'a' (opposite to angle A) is BC = 12 cm. The length of side 'b' (opposite to angle B) is AC = 13 cm. We need to find the third side AB using the Pythagorean theorem.
In \( \triangle ABC \), with \( \angle B = 90^\circ \):
\( AB^2 + BC^2 = AC^2 \)
\( AB^2 + 12^2 = 13^2 \)
\( AB^2 + 144 = 169 \)
\( AB^2 = 169 - 144 \)
\( AB^2 = 25 \)
\( AB = \sqrt{25} \)
\( AB = 5 \text{ cm} \)
Now we have all three sides: \( AB = 5 \text{ cm} \), \( BC = 12 \text{ cm} \), and \( AC = 13 \text{ cm} \). We can find the trigonometric ratios for angles A and C.

For acute angle A:
Opposite side to A is BC = 12 cm.
Adjacent side to A is AB = 5 cm.
Hypotenuse is AC = 13 cm.
\( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12}{13} \)
\( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5}{13} \)
\( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{12}{5} \)
\( \csc A = \frac{1}{\sin A} = \frac{13}{12} \)
\( \sec A = \frac{1}{\cos A} = \frac{13}{5} \)
\( \cot A = \frac{1}{\tan A} = \frac{5}{12} \)

For acute angle C:
Opposite side to C is AB = 5 cm.
Adjacent side to C is BC = 12 cm.
Hypotenuse is AC = 13 cm.
\( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5}{13} \)
\( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12}{13} \)
\( \tan C = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{5}{12} \)
\( \csc C = \frac{1}{\sin C} = \frac{13}{5} \)
\( \sec C = \frac{1}{\cos C} = \frac{13}{12} \)
\( \cot C = \frac{1}{\tan C} = \frac{12}{5} \)
In simple words: First, we used the Pythagoras formula to find the length of the third side. Then, for each angle (A and C), we identified the side opposite to it, the side next to it, and the longest side (hypotenuse). Using these, we wrote down all six trigonometric ratios.

🎯 Exam Tip: Remember that for complementary angles like A and C in a right-angled triangle, \( \sin A = \cos C \) and \( \cos A = \sin C \), which can be a quick check for your calculations.

 

Question 3. If \( \tan A = \sqrt{2} - 1 \), then prove that \( \sin A \cos A = \frac{1}{2\sqrt{2}} \).
Answer: We are given \( \tan A = \sqrt{2} - 1 \). We can write this as \( \tan A = \frac{\sqrt{2} - 1}{1} \).
In a right-angled triangle, \( \tan A = \frac{\text{Perpendicular}}{\text{Base}} \).
So, let the perpendicular (BC) be \( (\sqrt{2} - 1)k \) and the base (AB) be \( 1k \), where k is any positive constant. The Baudhayana formula (Pythagorean theorem) helps us find the hypotenuse (AC).
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (k)^2 + ((\sqrt{2} - 1)k)^2 \)
\( AC^2 = k^2 + (2 - 2\sqrt{2} + 1)k^2 \)
\( AC^2 = k^2 + (3 - 2\sqrt{2})k^2 \)
\( AC^2 = (1 + 3 - 2\sqrt{2})k^2 \)
\( AC^2 = (4 - 2\sqrt{2})k^2 \)
\( AC = \sqrt{(4 - 2\sqrt{2})k^2} \)
\( AC = k\sqrt{4 - 2\sqrt{2}} \)

Now, we find \( \sin A \) and \( \cos A \):
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{(\sqrt{2} - 1)k}{k\sqrt{4 - 2\sqrt{2}}} = \frac{\sqrt{2} - 1}{\sqrt{4 - 2\sqrt{2}}} \)
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k}{k\sqrt{4 - 2\sqrt{2}}} = \frac{1}{\sqrt{4 - 2\sqrt{2}}} \)

Next, we calculate \( \sin A \cos A \):
\( \sin A \cos A = \left(\frac{\sqrt{2} - 1}{\sqrt{4 - 2\sqrt{2}}}\right) \times \left(\frac{1}{\sqrt{4 - 2\sqrt{2}}}\right) \)
\( \sin A \cos A = \frac{\sqrt{2} - 1}{4 - 2\sqrt{2}} \)
To simplify, we can factor out \( \sqrt{2} \) from the numerator and \( 2 \) from the denominator:
\( \sin A \cos A = \frac{\sqrt{2} - 1}{2(2 - \sqrt{2})} \)
Factor out \( \sqrt{2} \) from the term \( (2 - \sqrt{2}) \):
\( \sin A \cos A = \frac{\sqrt{2} - 1}{2\sqrt{2}(\sqrt{2} - 1)} \)
Cancel out \( (\sqrt{2} - 1) \) from the numerator and denominator:
\( \sin A \cos A = \frac{1}{2\sqrt{2}} \)
This proves the given statement.
In simple words: We started with \( \tan A \) and used it to imagine a right-angled triangle. We found the lengths of all three sides using the Pythagoras rule. Then, we calculated \( \sin A \) and \( \cos A \). Finally, we multiplied \( \sin A \) and \( \cos A \) together and simplified the answer to show it matches \( \frac{1}{2\sqrt{2}} \).

🎯 Exam Tip: When dealing with expressions like \( \tan A = \sqrt{2} - 1 \), remember to rationalize the denominator or factor common terms carefully to simplify the final product of \( \sin A \cos A \).

 

Question 4. If \( \sin A = \frac{1}{3} \), then find the value of \( \cos A \cdot \csc A + \tan A \cdot \sec A \).
Answer: We are given \( \sin A = \frac{1}{3} \).
In a right-angled triangle, \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \).
So, let the perpendicular (BC) be \( 1k \) and the hypotenuse (AC) be \( 3k \), where k is any positive constant. We use the Baudhayana formula (Pythagorean theorem) to find the base (AB).
\( AB^2 + BC^2 = AC^2 \)
\( AB^2 + (k)^2 = (3k)^2 \)
\( AB^2 + k^2 = 9k^2 \)
\( AB^2 = 9k^2 - k^2 \)
\( AB^2 = 8k^2 \)
\( AB = \sqrt{8k^2} = 2\sqrt{2}k \)
Since A is an acute angle, AB must be positive.

Now we find the required trigonometric ratios:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{2\sqrt{2}k}{3k} = \frac{2\sqrt{2}}{3} \)
\( \csc A = \frac{1}{\sin A} = \frac{1}{1/3} = 3 \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{k}{2\sqrt{2}k} = \frac{1}{2\sqrt{2}} \)
\( \sec A = \frac{1}{\cos A} = \frac{1}{2\sqrt{2}/3} = \frac{3}{2\sqrt{2}} \)

Now, we find the value of the expression \( \cos A \cdot \csc A + \tan A \cdot \sec A \):
\( \cos A \cdot \csc A + \tan A \cdot \sec A = \left(\frac{2\sqrt{2}}{3}\right) \cdot (3) + \left(\frac{1}{2\sqrt{2}}\right) \cdot \left(\frac{3}{2\sqrt{2}}\right) \)
\( = 2\sqrt{2} + \frac{3}{(2\sqrt{2})(2\sqrt{2})} \)
\( = 2\sqrt{2} + \frac{3}{4 \times 2} \)
\( = 2\sqrt{2} + \frac{3}{8} \)
To combine these, find a common denominator:
\( = \frac{2\sqrt{2} \times 8}{8} + \frac{3}{8} \)
\( = \frac{16\sqrt{2} + 3}{8} \)
In simple words: First, we used the given \( \sin A \) value and the Pythagoras rule to find all sides of a right-angled triangle. Then, we calculated \( \cos A \), \( \csc A \), \( \tan A \), and \( \sec A \). Finally, we put these values into the given math expression and did the calculations to get the final answer.

🎯 Exam Tip: When you have a fraction with \( \sqrt{2} \) or other square roots in the denominator, remember to rationalize it by multiplying both the numerator and denominator by the square root to simplify the expression.

 

Question 5. If \( \cos \theta = \frac{8}{17} \), then find all the remaining trigonometrical ratios.
Answer: We are given \( \cos \theta = \frac{8}{17} \).
In a right-angled triangle, \( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \).
So, let the base (BC) be \( 8k \) and the hypotenuse (AC) be \( 17k \), where k is a positive constant. We use the Baudhayana formula (Pythagorean theorem) to find the perpendicular (AB).
\( AB^2 + BC^2 = AC^2 \)
\( AB^2 + (8k)^2 = (17k)^2 \)
\( AB^2 + 64k^2 = 289k^2 \)
\( AB^2 = 289k^2 - 64k^2 \)
\( AB^2 = 225k^2 \)
\( AB = \sqrt{225k^2} \)
\( AB = 15k \)
Since \( \theta \) is an acute angle, AB must be positive.

Now we have all three sides: \( AB = 15k \), \( BC = 8k \), and \( AC = 17k \). We can find the remaining trigonometric ratios.
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{15k}{17k} = \frac{15}{17} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} = \frac{15k}{8k} = \frac{15}{8} \)
\( \csc \theta = \frac{1}{\sin \theta} = \frac{17}{15} \)
\( \sec \theta = \frac{1}{\cos \theta} = \frac{17}{8} \)
\( \cot \theta = \frac{1}{\tan \theta} = \frac{8}{15} \)
In simple words: We used the given \( \cos \theta \) value and the Pythagoras rule to find the lengths of all three sides of a right-angled triangle. With these side lengths, we then calculated the values for \( \sin \theta \), \( \tan \theta \), \( \csc \theta \), \( \sec \theta \), and \( \cot \theta \).

🎯 Exam Tip: It's good practice to remember common Pythagorean triples (like 8, 15, 17) as this can speed up calculations for finding the third side of a right-angled triangle.

 

Question 6. If \( \cos A = \frac{5}{13} \), then find the value of \( \frac{\csc A}{\cos A + \csc A} \).
Answer: We are given \( \cos A = \frac{5}{13} \).
In a right-angled triangle, \( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} \).
So, let the base (AB) be \( 5k \) and the hypotenuse (AC) be \( 13k \), where k is a positive constant. We use the Baudhayana formula (Pythagorean theorem) to find the perpendicular (BC).
\( AB^2 + BC^2 = AC^2 \)
\( (5k)^2 + BC^2 = (13k)^2 \)
\( 25k^2 + BC^2 = 169k^2 \)
\( BC^2 = 169k^2 - 25k^2 \)
\( BC^2 = 144k^2 \)
\( BC = \sqrt{144k^2} \)
\( BC = 12k \)
Since A is an acute angle, BC must be positive.

Now we find the required trigonometric ratios for the expression:
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12k}{13k} = \frac{12}{13} \)
\( \csc A = \frac{1}{\sin A} = \frac{13}{12} \)

Now, we substitute these values into the given expression \( \frac{\csc A}{\cos A + \csc A} \):
\( = \frac{\frac{13}{12}}{\frac{5}{13} + \frac{13}{12}} \)
First, simplify the denominator:
\( \frac{5}{13} + \frac{13}{12} = \frac{5 \times 12 + 13 \times 13}{13 \times 12} = \frac{60 + 169}{156} = \frac{229}{156} \)
Now substitute this back into the expression:
\( = \frac{\frac{13}{12}}{\frac{229}{156}} \)
\( = \frac{13}{12} \times \frac{156}{229} \)
We know that \( 156 = 12 \times 13 \).
\( = \frac{13}{12} \times \frac{12 \times 13}{229} \)
Cancel out 12 from numerator and denominator:
\( = \frac{13 \times 13}{229} \)
\( = \frac{169}{229} \)
In simple words: We first used the given \( \cos A \) value and the Pythagoras theorem to find all the sides of the triangle. Then, we found \( \csc A \). Finally, we put all these values into the given math expression and solved it step-by-step.

🎯 Exam Tip: When dealing with complex fractions, it's often easiest to simplify the numerator and denominator separately before dividing. Also, look for opportunities to cancel common factors to make calculations simpler.

 

Question 7. If \( 5 \tan \theta = 4 \), then find the value of \( \frac{5\sin \theta - 3\cos \theta}{\sin \theta + 2\cos \theta} \).
Answer: We are given \( 5 \tan \theta = 4 \), which means \( \tan \theta = \frac{4}{5} \).

**Method I: Using a Right-Angled Triangle**
In a right-angled triangle, \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \).
Let the perpendicular (BC) be \( 4k \) and the base (AB) be \( 5k \), where k is a positive constant. We use the Baudhayana formula (Pythagorean theorem) to find the hypotenuse (AC).
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (5k)^2 + (4k)^2 \)
\( AC^2 = 25k^2 + 16k^2 \)
\( AC^2 = 41k^2 \)
\( AC = \sqrt{41k^2} = k\sqrt{41} \)

Now we find \( \sin \theta \) and \( \cos \theta \):
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4k}{k\sqrt{41}} = \frac{4}{\sqrt{41}} \)
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5k}{k\sqrt{41}} = \frac{5}{\sqrt{41}} \)

Now, we substitute these values into the given expression \( \frac{5\sin \theta - 3\cos \theta}{\sin \theta + 2\cos \theta} \):
\( = \frac{5\left(\frac{4}{\sqrt{41}}\right) - 3\left(\frac{5}{\sqrt{41}}\right)}{\left(\frac{4}{\sqrt{41}}\right) + 2\left(\frac{5}{\sqrt{41}}\right)} \)
\( = \frac{\frac{20}{\sqrt{41}} - \frac{15}{\sqrt{41}}}{\frac{4}{\sqrt{41}} + \frac{10}{\sqrt{41}}} \)
\( = \frac{\frac{20 - 15}{\sqrt{41}}}{\frac{4 + 10}{\sqrt{41}}} \)
\( = \frac{\frac{5}{\sqrt{41}}}{\frac{14}{\sqrt{41}}} \)
\( = \frac{5}{\sqrt{41}} \times \frac{\sqrt{41}}{14} \)
\( = \frac{5}{14} \)

**Method II: Direct Method (recommended for efficiency)**
We are given \( \tan \theta = \frac{4}{5} \).
The expression is \( \frac{5\sin \theta - 3\cos \theta}{\sin \theta + 2\cos \theta} \).
To simplify this expression using \( \tan \theta \), we can divide the numerator and the denominator by \( \cos \theta \).
\( = \frac{\frac{5\sin \theta}{\cos \theta} - \frac{3\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{2\cos \theta}{\cos \theta}} \)
We know that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \).
\( = \frac{5\tan \theta - 3}{\tan \theta + 2} \)
Now, substitute the value \( \tan \theta = \frac{4}{5} \):
\( = \frac{5\left(\frac{4}{5}\right) - 3}{\frac{4}{5} + 2} \)
\( = \frac{4 - 3}{\frac{4}{5} + \frac{10}{5}} \)
\( = \frac{1}{\frac{14}{5}} \)
\( = 1 \times \frac{5}{14} \)
\( = \frac{5}{14} \)
In simple words: We first found that \( \tan \theta \) is \( \frac{4}{5} \). Then, we changed the given math expression by dividing everything by \( \cos \theta \). This made the expression use \( \tan \theta \) instead of \( \sin \theta \) and \( \cos \theta \). Finally, we put the value of \( \tan \theta \) into the new expression to get the answer.

🎯 Exam Tip: For expressions involving \( \sin \theta \) and \( \cos \theta \) which need to be simplified given \( \tan \theta \), the direct method of dividing numerator and denominator by \( \cos \theta \) is usually faster and less prone to calculation errors than using a triangle to find individual \( \sin \theta \) and \( \cos \theta \) values.

 

Question 8. In a \( \triangle ABC \), \( \angle C = 90^\circ \) and if \( \cot A = \sqrt{3} \) and \( \cot B = \frac{1}{\sqrt{3}} \), then prove that \( \sin A \cos B + \cos A \sin B = 1 \).
Answer: We are given a right-angled triangle \( \triangle ABC \) where \( \angle C = 90^\circ \).

Given \( \cot A = \sqrt{3} \).
In a right-angled triangle, \( \cot A = \frac{\text{Base (adjacent to A)}}{\text{Perpendicular (opposite to A)}} = \frac{AC}{BC} \).
So, let \( AC = \sqrt{3}k \) and \( BC = k \), where k is a positive constant.
Using the Baudhayana theorem:
\( AB^2 = AC^2 + BC^2 \)
\( AB^2 = (\sqrt{3}k)^2 + (k)^2 \)
\( AB^2 = 3k^2 + k^2 \)
\( AB^2 = 4k^2 \)
\( AB = \sqrt{4k^2} = 2k \)

From these side lengths, we can find \( \sin A \) and \( \cos A \):
\( \sin A = \frac{BC}{AB} = \frac{k}{2k} = \frac{1}{2} \)
\( \cos A = \frac{AC}{AB} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \)

Given \( \cot B = \frac{1}{\sqrt{3}} \).
In the same right-angled triangle \( \triangle ABC \), \( \cot B = \frac{\text{Base (adjacent to B)}}{\text{Perpendicular (opposite to B)}} = \frac{BC}{AC} \).
This means \( BC = k' \) and \( AC = \sqrt{3}k' \). If we use the values from \( \cot A \), we have \( AC = \sqrt{3}k \) and \( BC = k \). These ratios are consistent.

Now we find \( \sin B \) and \( \cos B \) using \( AC = \sqrt{3}k \), \( BC = k \), and \( AB = 2k \):
\( \sin B = \frac{AC}{AB} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \)
\( \cos B = \frac{BC}{AB} = \frac{k}{2k} = \frac{1}{2} \)

Now, we need to prove \( \sin A \cos B + \cos A \sin B = 1 \).
Substitute the values we found:
L.H.S. \( = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{1}{4} + \frac{3}{4} \)
\( = \frac{1+3}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: We started by using the given \( \cot A \) value to find all the sides of the right-angled triangle using the Pythagoras rule. From these sides, we found \( \sin A \), \( \cos A \), \( \sin B \), and \( \cos B \). Then, we put these values into the expression \( \sin A \cos B + \cos A \sin B \) and showed that it equals 1. This means we proved the identity.

🎯 Exam Tip: This problem demonstrates the angle addition formula for sine: \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). Since \( \angle C = 90^\circ \), \( A+B = 90^\circ \), so \( \sin(A+B) = \sin(90^\circ) = 1 \), which provides a quick check for the result.

 

Question 10. In figure, AD = DB and \( \angle B = 90^\circ \) then find the value of the following:
(i) \( \sin \theta \)
(ii) \( \cos \theta \)
(iii) \( \tan \theta \)
Answer: We are given a right-angled triangle ABC with \( \angle B = 90^\circ \). Let AC = b and AB = a. Since AD = DB, D is the midpoint of AB. So, \( BD = \frac{AB}{2} = \frac{a}{2} \).
Using the Pythagorean theorem in \( \triangle ABC \):
\( AC^2 = AB^2 + BC^2 \)
\( b^2 = a^2 + BC^2 \)
\( BC^2 = b^2 - a^2 \)
\( BC = \sqrt{b^2 - a^2} \)

Now, consider the right-angled triangle \( \triangle BDC \) where \( \angle B = 90^\circ \). The angle \( \theta \) is \( \angle BDC \).
First, find the hypotenuse DC for \( \triangle BDC \):
\( DC^2 = BD^2 + BC^2 \)
\( DC^2 = \left(\frac{a}{2}\right)^2 + (\sqrt{b^2 - a^2})^2 \)
\( DC^2 = \frac{a^2}{4} + b^2 - a^2 \)
\( DC^2 = \frac{a^2 + 4b^2 - 4a^2}{4} \)
\( DC^2 = \frac{4b^2 - 3a^2}{4} \)
\( DC = \sqrt{\frac{4b^2 - 3a^2}{4}} \)
\( DC = \frac{\sqrt{4b^2 - 3a^2}}{2} \)

Now we can find the trigonometric ratios for \( \theta \):
(i) \( \sin \theta = \frac{\text{Opposite side to } \theta}{\text{Hypotenuse}} = \frac{BC}{DC} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{4b^2 - 3a^2}}{2}} = \frac{2\sqrt{b^2 - a^2}}{\sqrt{4b^2 - 3a^2}} \)
(ii) \( \cos \theta = \frac{\text{Adjacent side to } \theta}{\text{Hypotenuse}} = \frac{BD}{DC} = \frac{\frac{a}{2}}{\frac{\sqrt{4b^2 - 3a^2}}{2}} = \frac{a}{\sqrt{4b^2 - 3a^2}} \)
(iii) \( \tan \theta = \frac{\text{Opposite side to } \theta}{\text{Adjacent side to } \theta} = \frac{BC}{BD} = \frac{\sqrt{b^2 - a^2}}{\frac{a}{2}} = \frac{2\sqrt{b^2 - a^2}}{a} \)
In simple words: We used the given information to draw out the triangle and mark its sides with 'a' and 'b'. First, we found the length of side BC using Pythagoras' rule in the big triangle. Then, in the smaller triangle BDC, we found the length of DC, which is its longest side. After that, we used these side lengths to write down the sine, cosine, and tangent values for angle theta.

🎯 Exam Tip: When a problem involves multiple triangles or points on a side (like point D on AB), carefully break down the problem into smaller right-angled triangles and apply the Pythagorean theorem and trigonometric definitions to each part separately.

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