CBSE Class 10 Mathematics Trigonometry Assignment Set 07

Question 1. In a \( \Delta OMP \), right angled at M, if OM = 5 cm, MP = 12 cm, find all the t- ratios of \( \angle MOP \). (let \( \theta \))
Answer: In a right-angled triangle \( OMP \) where the right angle is at \( M \), we can find the hypotenuse \( OP \) by applying Pythagoras' theorem: \( OP^2 = OM^2 + MP^2 \)
\( \implies OP^2 = 5^2 + 12^2 = 25 + 144 = 169 \)
\( \implies OP = 13\text{ cm} \). For the angle \( \angle MOP = \theta \), the side roles are: - Base (adjacent) \( = OM = 5\text{ cm} \) - Perpendicular (opposite) \( = MP = 12\text{ cm} \) - Hypotenuse \( = OP = 13\text{ cm} \) The corresponding trigonometric ratios are calculated below: - \( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{12}{13} \) - \( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{5}{13} \) - \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{12}{5} \) - \( \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{5}{12} \) - \( \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{13}{5} \) - \( \csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{13}{12} \)
In simple words: First, find the longest side of the right triangle using Pythagoras' theorem, which is 13. Then, use the simple ratios of the sides to write down all six trigonometric values.
Exam Tip: Remember to clearly define which sides are the perpendicular (opposite) and base (adjacent) with respect to the angle \( \theta \) before writing down the ratios.

Question 2. In the adjoining \( \Delta ABC \) find all t- ratios of \( \angle C \) if AB = 4cm, BC = 5cm.
Answer: In the given right-angled triangle \( ABC \) (right-angled at \( A \)), we can use Pythagoras' theorem to find the length of the third side \( AC \): \( BC^2 = AB^2 + AC^2 \)
\( \implies 5^2 = 4^2 + AC^2 \)
\( \implies 25 = 16 + AC^2 \)
\( \implies AC^2 = 9 \)
\( \implies AC = 3\text{ cm} \). For the angle \( \angle C \), we define the sides as: - Perpendicular (opposite) \( = AB = 4\text{ cm} \) - Base (adjacent) \( = AC = 3\text{ cm} \) - Hypotenuse \( = BC = 5\text{ cm} \) The trigonometric ratios for \( \angle C \) are: - \( \sin C = \frac{AB}{BC} = \frac{4}{5} \) - \( \cos C = \frac{AC}{BC} = \frac{3}{5} \) - \( \tan C = \frac{AB}{AC} = \frac{4}{3} \) - \( \cot C = \frac{AC}{AB} = \frac{3}{4} \) - \( \sec C = \frac{BC}{AC} = \frac{5}{3} \) - \( \csc C = \frac{BC}{AB} = \frac{5}{4} \) In simple words: Find the third side of the triangle first, which is 3 cm. Then use the basic definitions of sine, cosine, tangent, and their reciprocals for angle C.

Exam Tip: Always state the right angle of the triangle clearly so that the hypotenuse is identified correctly as the side opposite to it.

Question 3. The diagonal PR and QS of a rhombus PQRS meet at O. If PR = 16cm, QS = 12cm. Find \( \cos \angle OPQ \).
Answer: Rhombus diagonals bisect each other perpendicularly at point \( O \). This gives: \( \angle POQ = 90^\circ \)
\( \implies OP = \frac{PR}{2} = \frac{16}{2} = 8\text{ cm} \)
\( \implies OQ = \frac{QS}{2} = \frac{12}{2} = 6\text{ cm} \). In the right-angled triangle \( POQ \), the hypotenuse \( PQ \) can be found as follows: \( PQ^2 = OP^2 + OQ^2 \)
\( \implies PQ^2 = 8^2 + 6^2 = 64 + 36 = 100 \)
\( \implies PQ = 10\text{ cm} \). For the angle \( \angle OPQ \) inside right-angled triangle \( POQ \): - Adjacent side (Base) \( = OP = 8\text{ cm} \) - Hypotenuse \( = PQ = 10\text{ cm} \) - \( \cos \angle OPQ = \frac{OP}{PQ} = \frac{8}{10} = \frac{4}{5} \). In simple words: The diagonals of a rhombus cut each other in half at 90 degrees. This creates a right-angled triangle inside, where we can find the outer side length and calculate the cosine value.

Exam Tip: Highlighting geometric properties like the perpendicular bisector nature of rhombus diagonals shows strong mathematical foundation and earns full step marks.

Question 4. If 3 Cot \( \theta \) = 4, find the value of \( \frac{3\sin\theta + 4\cos\theta}{3\sin\theta - 2\cos\theta} \) without using triangle method
Answer: We are given: \( 3 \cot \theta = 4 \)
\( \implies \cot \theta = \frac{4}{3} \). We need to evaluate the expression: \( \frac{3 \sin \theta + 4 \cos \theta}{3 \sin \theta - 2 \cos \theta} \) Dividing both the numerator and the denominator of the fraction by \( \sin \theta \): \( \frac{\frac{3\sin\theta + 4\cos\theta}{\sin\theta}}{\frac{3\sin\theta - 2\cos\theta}{\sin\theta}} = \frac{3 + 4\cot\theta}{3 - 2\cot\theta} \) By substituting the given value of \( \cot \theta = \frac{4}{3} \): \( \frac{3 + 4\left(\frac{4}{3}\right)}{3 - 2\left(\frac{4}{3}\right)} = \frac{3 + \frac{16}{3}}{3 - \frac{8}{3}} = \frac{\frac{9+16}{3}}{\frac{9-8}{3}} = \frac{25}{1} = 25 \).
In simple words: Divide the top and bottom of the fraction by sine to turn them into cotangent terms. This allows you to plug in the given cotangent value directly.

Exam Tip: Always read the instructions carefully - when "without using triangle method" is specified, using a triangle to find individual values of sine and cosine will lead to a penalty.

Question 5. In \( \Delta ABC \), right angled at B. If AC + BC = 25 cm. and AB = 5 cm. Find all t- ratios of \( \angle ACB \)
Answer: Let the side \( BC = x \). Since \( AC + BC = 25\text{ cm} \): \( AC = 25 - x \). In right-angled triangle \( ABC \) (right-angled at \( B \)), using Pythagoras' theorem: \( AC^2 = AB^2 + BC^2 \)
\( \implies (25 - x)^2 = 5^2 + x^2 \)
\( \implies 625 - 50x + x^2 = 25 + x^2 \)
\( \implies 600 = 50x \)
\( \implies x = 12\text{ cm} \). So the side lengths are: - \( BC = 12\text{ cm} \) - \( AC = 25 - 12 = 13\text{ cm} \) - \( AB = 5\text{ cm} \) For \( \angle ACB = C \): - Perpendicular (opposite) \( = AB = 5\text{ cm} \) - Base (adjacent) \( = BC = 12\text{ cm} \) - Hypotenuse \( = AC = 13\text{ cm} \) The corresponding trigonometric ratios of \( \angle ACB \) are: - \( \sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13} \) - \( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{13} \) - \( \tan C = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12} \) - \( \cot C = \frac{\text{Base}}{\text{Perpendicular}} = \frac{12}{5} \) - \( \sec C = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{13}{12} \) - \( \csc C = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{13}{5} \)
In simple words: Use the sum of the sides and Pythagoras' theorem to set up an equation. Solve it to find the three side lengths (5, 12, and 13) and write down the ratios.

Exam Tip: Be careful when expanding algebraic terms like \( (25-x)^2 \); mistakes in identity expansion are a common source of lost marks.

Question 6. In \( \Delta ABC \), right angled at A. If AB = 7cm., AC = 24cm. Find the value of \( \sin B\cos C + \cos B\sin C \).
Answer: We can find the hypotenuse \( BC \) using Pythagoras' theorem: \( BC^2 = AB^2 + AC^2 \)
\( \implies BC^2 = 7^2 + 24^2 = 49 + 576 = 625 \)
\( \implies BC = 25\text{ cm} \). Next, determine the values for angles \( B \) and \( C \): - For angle \( B \): - \( \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{24}{25} \) - \( \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{BC} = \frac{7}{25} \) - For angle \( C \): - \( \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{BC} = \frac{7}{25} \) - \( \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{24}{25} \) Substituting these values into our expression gives: \( \sin B \cos C + \cos B \sin C = \left(\frac{24}{25}\right)\left(\frac{24}{25}\right) + \left(\frac{7}{25}\right)\left(\frac{7}{25}\right) \)
\( \implies \frac{576}{625} + \frac{49}{625} = \frac{625}{625} = 1 \).
In simple words: Find the longest side first (25 cm). Then, plug the sine and cosine values of both angles into the formula to calculate the final answer, which simplifies to 1.

Exam Tip: Alternatively, you can verify this using the identity \( \sin(B+C) \). Since \( \angle A = 90^\circ \), \( B+C = 90^\circ \), meaning \( \sin(B+C) = \sin(90^\circ) = 1 \).

Question 7. In a \( \Delta ABC \), right angled at A. If \( \tan B = \frac{1}{\sqrt{3}} \) and \( \tan C = \sqrt{3} \), Show that \( \cos B\cos C + \sin B\sin C = \frac{\sqrt{3}}{2} \)
Answer: From standard angles: \( \tan B = \frac{1}{\sqrt{3}} \)
\( \implies B = 30^\circ \) And: \( \tan C = \sqrt{3} \)
\( \implies C = 60^\circ \). Substituting these angles into the expression: \( \cos B \cos C + \sin B \sin C = \cos 30^\circ \cos 60^\circ + \sin 30^\circ \sin 60^\circ \)
\( \implies \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \)
\( \implies \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \). The LHS is equal to the RHS, verifying the equation.
In simple words: Find the angles using the tangent values (30 and 60 degrees). Substitute these angles into the expression to prove the given equation.

Exam Tip: Recognizing standard angle values instantly saves calculation time during exams.

Question 8. If Cot A = \( \frac{12}{5} \), Verify that \( \tan^2 A - \sin^2 A = \sin^4 A\sec^2 A \)
Answer: Given: \( \cot A = \frac{12}{5} \)
\( \implies \text{Base} = 12 \text{ and Perpendicular} = 5 \). We find the hypotenuse as follows: \( \text{Hypotenuse} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \). Hence, the remaining ratios are: - \( \tan A = \frac{5}{12} \) - \( \sin A = \frac{5}{13} \) - \( \sec A = \frac{13}{12} \) Let us evaluate both sides: LHS \( = \tan^2 A - \sin^2 A \)
\( \implies \left(\frac{5}{12}\right)^2 - \left(\frac{5}{13}\right)^2 = \frac{25}{144} - \frac{25}{169} \)
\( \implies \frac{25(169 - 144)}{144 \times 169} = \frac{25 \times 25}{24336} = \frac{625}{24336} \). RHS \( = \sin^4 A \sec^2 A \)
\( \implies \left(\frac{5}{13}\right)^4 \left(\frac{13}{12}\right)^2 = \frac{625}{28561} \times \frac{169}{144} = \frac{625}{169 \times 144} = \frac{625}{24336} \). Since LHS = RHS, the identity is verified.
In simple words: Find the three side lengths using the cotangent value. Substitute these values into both sides of the equation to show they are equal.

Exam Tip: You can also prove this identity algebraically first by rewriting tangent as sine/cosine, which simplifies the verification process.

Question 9. In \( \Delta PQR \), right angled at P. If 4 Sec Q - 3 = 5, Find tan R
Answer: Starting with: \( 4 \sec Q - 3 = 5 \)
\( \implies 4 \sec Q = 8 \)
\( \implies \sec Q = 2 \). Since the angle \( Q \) is acute and \( \sec 60^\circ = 2 \), we get: \( Q = 60^\circ \). In right-angled triangle \( PQR \) with \( \angle P = 90^\circ \): \( \angle P + \angle Q + \angle R = 180^\circ \)
\( \implies 90^\circ + 60^\circ + R = 180^\circ \)
\( \implies R = 30^\circ \). Thus, we find the value of \( \tan R \): \( \tan R = \tan 30^\circ = \frac{1}{\sqrt{3}} \).
In simple words: Solve the equation to find that angle Q is 60 degrees. Since the other angle is 90 degrees, angle R must be 30 degrees. The tangent of 30 degrees is \( 1/\sqrt{3} \).

Exam Tip: For triangle problems involving standard ratios, using angle values directly is often much faster and less prone to arithmetic error than working with variable side lengths.

Question 10. If tan \( \theta \) = \( \frac{a}{b} \), Without using the triangle, Find the Value of \( \frac{a\sin\theta - b\cos\theta}{a\sin\theta + b\cos\theta} \)
Answer: We have: \( \tan \theta = \frac{a}{b} \) To find the value of the given expression: \( \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} \) We can divide both the numerator and the denominator by \( \cos \theta \): \( \frac{\frac{a\sin\theta - b\cos\theta}{\cos\theta}}{\frac{a\sin\theta + b\cos\theta}{\cos\theta}} = \frac{a\tan\theta - b}{a\tan\theta + b} \) Now substitute \( \tan \theta = \frac{a}{b} \):
\( \implies \frac{a\left(\frac{a}{b}\right) - b}{a\left(\frac{a}{b}\right) + b} = \frac{\frac{a^2}{b} - b}{\frac{a^2}{b} + b} = \frac{\frac{a^2 - b^2}{b}}{\frac{a^2 + b^2}{b}} = \frac{a^2 - b^2}{a^2 + b^2} \).
In simple words: Divide everything by cosine to turn sine into tangent. Then replace tangent with the fraction \( a/b \) and simplify the expression.

Exam Tip: When simplifying complex fractions, make sure to find a common denominator for the top and bottom terms before canceling them out.

Question 11. From the adjoining figure, find the value of
(i) \( \frac{\tan \angle ACB}{\tan \angle ADB} \)
(ii) \( \frac{\tan \angle CAB}{\tan \angle DAB} \)
Answer: In the right-angled triangle \( ABD \) (with the right angle at \( B \)), the point \( C \) on the base \( BD \) acts as the midpoint, meaning: \( BD = 2 BC \). Let's find each ratio: (i) For the first ratio: - In triangle \( ABC \): \( \tan \angle ACB = \frac{AB}{BC} \) - In triangle \( ABD \): \( \tan \angle ADB = \frac{AB}{BD} \) Thus: \( \frac{\tan \angle ACB}{\tan \angle ADB} = \frac{\frac{AB}{BC}}{\frac{AB}{BD}} = \frac{BD}{BC} = \frac{2 BC}{BC} = 2 \). (ii) For the second ratio: - In triangle \( ABC \): \( \tan \angle CAB = \frac{BC}{AB} \) - In triangle \( ABD \): \( \tan \angle DAB = \frac{BD}{AB} \) Thus: \( \frac{\tan \angle CAB}{\tan \angle DAB} = \frac{\frac{BC}{AB}}{\frac{BD}{AB}} = \frac{BC}{BD} = \frac{BC}{2 BC} = \frac{1}{2} \). In simple words: Use the basic tangent formula (opposite over adjacent) for both triangles. Since C is right in the middle of BD, the base of the larger triangle is exactly twice as long, giving ratios of 2 and 1/2.

Exam Tip: In problems where no numbers are given on the sides, expressing ratios in terms of a common side segment (like AB or BC) helps simplify the final fraction easily.

Question 12. From the adjoining figure, Find the Value "Cot \( \theta \)"
Answer: Let us construct a perpendicular line segment \( RT \) from \( R \) to the line \( PS \). This construction yields a rectangle \( PQRT \) and a right-angled triangle \( RTS \) at \( T \). In the rectangle \( PQRT \): - \( PT = RQ = 5\text{ cm} \) - \( RT = PQ \) Using the length of \( PS = 14\text{ cm} \): \( ST = PS - PT = 14 - 5 = 9\text{ cm} \). In the right-angled triangle \( PQR \) (right-angled at \( Q \)): \( PQ^2 = PR^2 - RQ^2 \)
\( \implies PQ^2 = 13^2 - 5^2 = 169 - 25 = 144 \)
\( \implies PQ = 12\text{ cm} \). Since \( RT = PQ \), we have \( RT = 12\text{ cm} \). In right triangle \( RTS \): \( \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{ST}{RT} = \frac{9}{12} = \frac{3}{4} \). *(Note: If the required ratio is tangent, then \( \tan \theta = \frac{RT}{ST} = \frac{12}{9} = \frac{4}{3} \).)* In simple words: Draw a line to split the shape into a rectangle and a right triangle. Find the missing sides, then use the cotangent ratio (adjacent side divided by opposite side) to get the answer.

Exam Tip: Draw clear auxiliary lines (like RT) on your main diagram and state your construction steps clearly to secure full marks.

Question 13. From the adjoining figure Find the value of \( (\tan A - \sin B + \cos A) \)
Answer: In the right-angled triangle \( COB \) (with the right angle at \( O \)): By Pythagoras' theorem: \( BC^2 = CO^2 + OB^2 \)
\( \implies BC^2 = 12^2 + 9^2 = 144 + 81 = 225 \)
\( \implies BC = 15\text{ cm} \). Next, consider right-angled triangle \( COA \): - Case 1: If the triangle has side lengths \( CO = 15\text{ cm} \), \( AO = 36\text{ cm} \), and \( AC = 39\text{ cm} \) (consistent with a standard 15-36-39 integer right triangle): - \( \tan A = \frac{CO}{AO} = \frac{15}{36} = \frac{5}{12} \) - \( \cos A = \frac{AO}{AC} = \frac{36}{39} = \frac{12}{13} \) - \( \sin B = \frac{CO}{BC} = \frac{12}{13} \) - \( \tan A - \sin B + \cos A = \frac{5}{12} - \frac{12}{13} + \frac{12}{13} = \frac{5}{12} \). - Case 2: If the labeled dimensions \( CO = 12\text{ cm} \), \( OB = 9\text{ cm} \), and \( AO = 39\text{ cm} \) are evaluated directly: - \( BC = 15\text{ cm} \) - \( AC = \sqrt{12^2 + 39^2} = \sqrt{1665}\text{ cm} \) - \( \tan A = \frac{CO}{AO} = \frac{12}{39} = \frac{4}{13} \) - \( \sin B = \frac{CO}{BC} = \frac{12}{15} = \frac{4}{5} \) - \( \cos A = \frac{AO}{AC} = \frac{39}{\sqrt{1665}} \) - \( \tan A - \sin B + \cos A = \frac{4}{13} - \frac{4}{5} + \frac{39}{\sqrt{1665}} \). In simple words: Find the hypotenuse for the right-hand triangle using Pythagoras' theorem to get 15. Then calculate each of the three trig values and substitute them to solve.

Exam Tip: When solving questions with potential printing or drafting discrepancies, write down your assumptions clearly so the examiner can follow your steps.

Question 14. If tan A = \( \sqrt{2} - 1 \) Show that Sin A . Cos A = \( \frac{\sqrt{2}}{4} \)
Answer: We can use the trigonometric identity: \( \sin A \cos A = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A} = \frac{\tan A}{\tan^2 A + 1} \) Substitute \( \tan A = \sqrt{2} - 1 \) into the identity: \( \sin A \cos A = \frac{\sqrt{2} - 1}{(\sqrt{2} - 1)^2 + 1} \)
\( \implies \frac{\sqrt{2} - 1}{(2 - 2\sqrt{2} + 1) + 1} \)
\( \implies \frac{\sqrt{2} - 1}{4 - 2\sqrt{2}} \) Factor out \( 2\sqrt{2} \) from the denominator: \( \frac{\sqrt{2} - 1}{2\sqrt{2}(\sqrt{2} - 1)} = \frac{1}{2\sqrt{2}} \) Rationalizing the denominator gives: \( \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4} \). The LHS is equal to the RHS, completing the proof.
In simple words: Express sine times cosine in terms of tangent to avoid working with complex square root triangles. Substituting the value of tangent and simplifying gives the exact required proof.

Exam Tip: Rationalization is a crucial step when working with surds in the denominator. Always multiply by the conjugate or root to simplify terms.

Question 15. In the adjoining figure Find the value of
(i) Sin B
(ii) tan C
(iii) \( \tan^2 B + \tan^2 C \)
(iv) \( \sin^2 C + \cos^2 C + \cos B \)
Answer: In the right-angled triangle \( ADB \) (right-angled at \( D \)): \( AD^2 = AB^2 - BD^2 \)
\( \implies AD^2 = 13^2 - 5^2 = 169 - 25 = 144 \)
\( \implies AD = 12\text{ cm} \). In the right-angled triangle \( ADC \) (right-angled at \( D \)): \( AC^2 = AD^2 + DC^2 \)
\( \implies AC^2 = 12^2 + 16^2 = 144 + 256 = 400 \)
\( \implies AC = 20\text{ cm} \). Now we compute each sub-part: (i) For \( \sin B \) in triangle \( ADB \): \( \sin B = \frac{AD}{AB} = \frac{12}{13} \). (ii) For \( \tan C \) in triangle \( ADC \): \( \tan C = \frac{AD}{DC} = \frac{12}{16} = \frac{3}{4} \). (iii) For \( \tan^2 B + \tan^2 C \): Since \( \tan B = \frac{12}{5} \) and \( \tan C = \frac{3}{4} \): \( \tan^2 B + \tan^2 C = \left(\frac{12}{5}\right)^2 + \left(\frac{3}{4}\right)^2 \)
\( \implies \frac{144}{25} + \frac{9}{16} = \frac{144 \times 16 + 9 \times 25}{400} = \frac{2304 + 225}{400} = \frac{2529}{400} \). (iv) For \( \sin^2 C + \cos^2 C + \cos B \): Using the fundamental identity \( \sin^2 C + \cos^2 C = 1 \): \( 1 + \cos B \) Since \( \cos B = \frac{BD}{AB} = \frac{5}{13} \): \( 1 + \frac{5}{13} = \frac{18}{13} \). In simple words: First find the common height (12) and the other hypotenuse (20) using Pythagoras' theorem. Then, write down the required values and simplify the expressions.

Exam Tip: Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) instantly simplifies complex-looking expressions like in sub-part (iv), reducing work and avoiding arithmetic errors.