Question: User: AAROHAN WORKSHEET 2 CLASS X 1. Assertion (A): If $\sin X + \cos Y = 1$; $X = 30^\circ$ AND $Y$ IS ACUTE ANGLE THEN $Y = 60^\circ$ Reason (R): $\sin 30^\circ = \frac{1}{2}$ AND $\cos 60^\circ = \frac{1}{2}$ 2. Assertion (A): The value of $\sin\theta = \frac{4}{3}$ is not possible Reason (R): Hypotenuse is the longest side in any right angle triangle. 3. Assertion (A): If $\tan\theta = \sin 45^\circ \cdot \cos 45^\circ + \sin 30^\circ$ then $\theta = 45^\circ$ Reason (R): $\tan 45^\circ = 1$ 4. Assertion (A): If $\triangle ABC$ is right angled at $C$ then the value of $\cos (A + B) = 0$ Reason (R): In $\triangle ABC$, $\angle A + \angle B + \angle C = 180^\circ$ 5. ASSERTION: $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$ REASON: $\sin x = \cos x$ for all values of $x$ 6. Show that: $\sec^2 \theta - \sin^2 \theta - 2 \tan^2 \theta = 1$ Only draw fig. for following 7. From the top of hill, the angles of depression of two consecutive kilometer stones due east are found to be $30^\circ$ and $45^\circ$ respectively. Find the height of the hill. 8. Two boats approach a right house in mid sea from opposite directions. The angles of elevation of the top of the lighthouse from two boats are $30^\circ$ and $45^\circ$ respectively. If the distance between two boats is 100m, find the height of the light house. 9. The angle of elevation of the top of a tower from a point is $30^\circ$. If the observer moves 20 metre towards the tower the angle of elevation of the top increases by $15^\circ$. Find the height of the tower. 10. An electrician needs to repair a lamp on a wall whose height is 7m. The lamp is 1 m below the top most point. She uses a ladder to reach the lamp as shown in the figure below. 1. What is the length of ladder? 2. How far is the foot of ladder from the wall? 3. What is the angle the ladder makes with the wall? Graph/figure: bottom-right, a ladder is leaning from the ground up to a point on a vertical wall; the wall is drawn as a dark vertical line on the right and the ladder as a diagonal line rising from bottom-left to mid-right.ans of all

1. **Problem:** Verify the assertion and reason for $\sin X + \cos Y = 1$ with $X=30^\circ$ and $Y$ acute. 2. **Problem:** Explain why $\sin\theta = \frac{4}{3}$ is not possible. 3. **Problem:** Given $\tan\theta = \sin 45^\circ \cdot \cos 45^\circ + \sin 30^\circ$, find $\theta$. 4. **Problem:** Show that if $\triangle ABC$ is right angled at $C$, then $\cos (A+B) = 0$. 5. **Problem:** Verify $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$ and discuss if $\sin x = \cos x$ for all $x$. 6. **Problem:** Prove $\sec^2 \theta - \sin^2 \theta - 2 \tan^2 \theta = 1$. 7. **Problem:** Find height of hill given angles of depression $30^\circ$ and $45^\circ$ to two stones 1 km apart. 8. **Problem:** Find height of lighthouse given angles of elevation $30^\circ$ and $45^\circ$ from two boats 100m apart. 9. **Problem:** Find height of tower given angle of elevation $30^\circ$, observer moves 20m closer, angle increases by $15^\circ$. 10. **Problem:** Ladder problem: wall height 7m, lamp 1m below top, find ladder length, foot distance, and angle. --- ### Step 1: Assertion 1 - Given $\sin X + \cos Y = 1$, $X=30^\circ$, $Y$ acute. - Calculate $\sin 30^\circ = \frac{1}{2}$. - Then $\cos Y = 1 - \sin 30^\circ = 1 - \frac{1}{2} = \frac{1}{2}$. - Since $\cos Y = \frac{1}{2}$ and $Y$ acute, $Y = 60^\circ$. ### Step 2: Assertion 2 - $\sin \theta$ ranges between $-1$ and $1$. - $\sin \theta = \frac{4}{3} > 1$ is impossible. - Reason: Hypotenuse is longest side, so sine (opposite/hypotenuse) cannot exceed 1. ### Step 3: Assertion 3 - Calculate $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$. - Compute $\sin 45^\circ \cdot \cos 45^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{1}{2}$. - Sum: $\frac{1}{2} + \frac{1}{2} = 1$. - So $\tan \theta = 1$ implies $\theta = 45^\circ$. ### Step 4: Assertion 4 - In $\triangle ABC$, $\angle A + \angle B + \angle C = 180^\circ$. - If right angled at $C$, $\angle C = 90^\circ$. - Then $\angle A + \angle B = 90^\circ$. - $\cos (A + B) = \cos 90^\circ = 0$. ### Step 5: Assertion 5 - $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$ is true. - But $\sin x = \cos x$ is not true for all $x$, only when $x = 45^\circ + k \times 180^\circ$. ### Step 6: Prove $\sec^2 \theta - \sin^2 \theta - 2 \tan^2 \theta = 1$ - Recall identities: - $\sec^2 \theta = 1 + \tan^2 \theta$ - $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$ - Substitute: $$\sec^2 \theta - \sin^2 \theta - 2 \tan^2 \theta = (1 + \tan^2 \theta) - \sin^2 \theta - 2 \tan^2 \theta = 1 - \sin^2 \theta - \tan^2 \theta$$ - Write $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$: $$1 - \sin^2 \theta - \frac{\sin^2 \theta}{\cos^2 \theta} = 1 - \sin^2 \theta \left(1 + \frac{1}{\cos^2 \theta}\right) = 1 - \sin^2 \theta \frac{\cos^2 \theta + 1}{\cos^2 \theta}$$ - Since $\cos^2 \theta + \sin^2 \theta = 1$, so $\cos^2 \theta + 1 = 1 + \cos^2 \theta$ (no direct simplification), but let's rewrite: $$= 1 - \sin^2 \theta \frac{1 + \cos^2 \theta}{\cos^2 \theta}$$ - This expression is not identically 1 for all $\theta$, so the original identity is false unless there's a typo. ### Step 7: Height of hill - Let height be $h$. - Let distances to stones be $d$ and $d+1$ km. - Angles of depression: $30^\circ$ and $45^\circ$. - From top, angle of depression equals angle of elevation from stones. - For stone at $d$ km: $$\tan 30^\circ = \frac{h}{d} \Rightarrow h = d \tan 30^\circ = d \times \frac{1}{\sqrt{3}}$$ - For stone at $d+1$ km: $$\tan 45^\circ = \frac{h}{d+1} = 1 \Rightarrow h = d + 1$$ - Equate: $$d \times \frac{1}{\sqrt{3}} = d + 1$$ - Multiply both sides by $\sqrt{3}$: $$d = (d + 1) \sqrt{3}$$ - Rearrange: $$d - d \sqrt{3} = \sqrt{3}$$ $$d(1 - \sqrt{3}) = \sqrt{3}$$ $$d = \frac{\sqrt{3}}{1 - \sqrt{3}} = \frac{\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{\sqrt{3} + 3}{1 - 3} = \frac{\sqrt{3} + 3}{-2}$$ - Negative distance is not possible, so switch order: $$d + 1 = h$$ $$h = d \times \frac{1}{\sqrt{3}}$$ - Substitute $h$: $$d + 1 = d \times \frac{1}{\sqrt{3}}$$ - Rearranged: $$d \times \frac{1}{\sqrt{3}} - d = 1$$ $$d \left(\frac{1}{\sqrt{3}} - 1\right) = 1$$ $$d = \frac{1}{\frac{1}{\sqrt{3}} - 1} = \frac{1}{\frac{1 - \sqrt{3}}{\sqrt{3}}} = \frac{\sqrt{3}}{1 - \sqrt{3}} = -\frac{\sqrt{3}}{\sqrt{3} - 1}$$ - Multiply numerator and denominator by $\sqrt{3} + 1$: $$d = -\frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = -\frac{3 + \sqrt{3}}{3 - 1} = -\frac{3 + \sqrt{3}}{2}$$ - Negative distance again, so the first stone is closer, so assign: $$d = h \tan 30^\circ = \frac{h}{\sqrt{3}}$$ $$d + 1 = h \tan 45^\circ = h$$ - So: $$d + 1 = h$$ $$d = \frac{h}{\sqrt{3}}$$ - Substitute $d$: $$\frac{h}{\sqrt{3}} + 1 = h$$ - Rearrange: $$1 = h - \frac{h}{\sqrt{3}} = h \left(1 - \frac{1}{\sqrt{3}}\right) = h \frac{\sqrt{3} - 1}{\sqrt{3}}$$ - Solve for $h$: $$h = \frac{1 \times \sqrt{3}}{\sqrt{3} - 1} = \frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + \sqrt{3}}{2} \approx 2.366 \text{ km}$$ ### Step 8: Height of lighthouse - Let height be $h$. - Distance between boats is 100m. - Let distances from boats to lighthouse base be $x$ and $100 - x$. - Angles of elevation: $30^\circ$ and $45^\circ$. - From boat 1: $$\tan 30^\circ = \frac{h}{x} = \frac{1}{\sqrt{3}} \Rightarrow h = \frac{x}{\sqrt{3}}$$ - From boat 2: $$\tan 45^\circ = \frac{h}{100 - x} = 1 \Rightarrow h = 100 - x$$ - Equate: $$\frac{x}{\sqrt{3}} = 100 - x$$ - Multiply both sides by $\sqrt{3}$: $$x = (100 - x) \sqrt{3}$$ - Rearrange: $$x + x \sqrt{3} = 100 \sqrt{3}$$ $$x (1 + \sqrt{3}) = 100 \sqrt{3}$$ $$x = \frac{100 \sqrt{3}}{1 + \sqrt{3}} = \frac{100 \sqrt{3} (1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{100 \sqrt{3} (1 - \sqrt{3})}{1 - 3} = \frac{100 \sqrt{3} (1 - \sqrt{3})}{-2}$$ - Calculate numerator: $$100 \sqrt{3} (1 - \sqrt{3}) = 100 \sqrt{3} - 100 \times 3 = 100 \sqrt{3} - 300$$ - So: $$x = -\frac{100 \sqrt{3} - 300}{2} = \frac{300 - 100 \sqrt{3}}{2}$$ - Approximate: $$\sqrt{3} \approx 1.732$$ $$x \approx \frac{300 - 173.2}{2} = \frac{126.8}{2} = 63.4 \text{ m}$$ - Height: $$h = \frac{x}{\sqrt{3}} = \frac{63.4}{1.732} \approx 36.6 \text{ m}$$ ### Step 9: Height of tower - Let height be $h$. - Initial distance from tower base: $d$. - Initial angle: $30^\circ$. - After moving 20m closer: distance $d - 20$. - New angle: $45^\circ$. - From initial position: $$\tan 30^\circ = \frac{h}{d} = \frac{1}{\sqrt{3}} \Rightarrow h = \frac{d}{\sqrt{3}}$$ - From new position: $$\tan 45^\circ = \frac{h}{d - 20} = 1 \Rightarrow h = d - 20$$ - Equate: $$\frac{d}{\sqrt{3}} = d - 20$$ - Rearrange: $$d - \frac{d}{\sqrt{3}} = 20$$ $$d \left(1 - \frac{1}{\sqrt{3}}\right) = 20$$ $$d = \frac{20}{1 - \frac{1}{\sqrt{3}}} = \frac{20}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = 20 \times \frac{\sqrt{3}}{\sqrt{3} - 1} = \frac{20 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20 \sqrt{3} (\sqrt{3} + 1)}{2} = 10 \sqrt{3} (\sqrt{3} + 1)$$ - Calculate: $$10 \sqrt{3} \times \sqrt{3} = 10 \times 3 = 30$$ $$10 \sqrt{3} \times 1 = 10 \sqrt{3} \approx 17.32$$ - So: $$d = 30 + 17.32 = 47.32 \text{ m}$$ - Height: $$h = d - 20 = 47.32 - 20 = 27.32 \text{ m}$$ ### Step 10: Ladder problem - Wall height: 7m - Lamp height: $7 - 1 = 6$ m - Let ladder length be $L$, distance from wall be $x$, angle with wall be $\theta$. 1. Length of ladder: $$L = \frac{6}{\sin \theta}$$ 2. Distance from wall: $$x = L \cos \theta = \frac{6}{\sin \theta} \cos \theta = 6 \cot \theta$$ 3. Angle $\theta$: - Using Pythagoras: $$L^2 = x^2 + 6^2$$ - Substitute $x = 6 \cot \theta$, $L = \frac{6}{\sin \theta}$: $$\left(\frac{6}{\sin \theta}\right)^2 = (6 \cot \theta)^2 + 6^2$$ $$\frac{36}{\sin^2 \theta} = 36 \cot^2 \theta + 36$$ - Divide both sides by 36: $$\frac{1}{\sin^2 \theta} = \cot^2 \theta + 1$$ - Recall identity: $$\cot^2 \theta + 1 = \csc^2 \theta = \frac{1}{\sin^2 \theta}$$ - Identity holds for all $\theta$, so ladder length and distance depend on chosen $\theta$. - If angle is given or ladder length is known, values can be computed. --- **Final answers:** 1. $Y = 60^\circ$. 2. $\sin \theta = \frac{4}{3}$ impossible. 3. $\theta = 45^\circ$. 4. $\cos (A+B) = 0$. 5. $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, but $\sin x = \cos x$ not always true. 6. Identity in question 6 is false as stated. 7. Height of hill $\approx 2.366$ km. 8. Height of lighthouse $\approx 36.6$ m. 9. Height of tower $\approx 27.32$ m. 10. Ladder length and foot distance depend on angle; lamp height is 6 m.