This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In ∆ABC, ∠C = 90°, ∠ABC =θ°, BC = 21 units and Ab = 29 unitsShow that : cos2– sin2 =\(\frac{41}{841}.\) |
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Answer» Given that AB = 29 units & BC = 21 units and ∠ABC =θ°& ∠ACB = 90°. Using Pythagoras theorem in right angled triangle ∆ACB, AB2 = AC2 + BC2 ⇒ AC2 = AB2 − BC2 ⇒ AC =\(\sqrt{AB^2-BC^2}\)=\(\sqrt{29^2-21^2}\)=\(\sqrt{841-441}=\sqrt{400}\) =20 units. (∵ BC = 21, AB = 29) Now, cosθ =\(\frac{BC}{AB}=\frac{21}{29}.\)(∵ BC = 21, AB = 29) And sinθ =\(\frac{AC}{AB}=\frac{20}{29}.\)(∵ AC = 20 and AB = 29) Now, cos2θ− sin2θ =\((\frac{21}{29})^2-(\frac{20}{29})^2\)=\(\frac{21^2-20^2}{29^2}\) =\(\frac{441 \,- \,400}{841}=\frac{41}{841}.\). Hence Proved |
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| 2. |
A student has set up a solar cooker in a box using a black painted aluminium sheet, a black cooking vessel, some glass wool , a transparent glass sheet and a mirror plate. What is the role of black paint, transparent glass sheet and a mirror used in the solar cooker ? |
| Answer» Black paint absorbs the heat energy quickly and thus helps in increasing the inner temperature. Glass sheet causes green-houses effect to retain the reflected infrared rays of heat, within the solar cooker. Mirror reflects the solar rays so that these rays reach inside the solar cooker. | |