Trigonometry Important Questions | Class 10 Maths | Q5

Chapter 8 Extra Questions | Trigonometric Ratios

This CBSE class 10 Maths extra question is from the topic Trigonometry. Concept Tested: computing trigonometric values for select angles and using that to find values of trigonometric expressions.

Question 5 : In triangle ABC, right angled at B if sin A = \\frac{\text{1}}{\text{2}}). Find the value of
1. sin C cos A – cos C sin A
2. cos A cos C + sin A sin C


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Explanatory Answer | Trigonometry Important Question 5

CBSE Class 10 Trigonometry Q8 Triangle

sin A = \\frac{\text{side opposite to ∠A}}{\text{hypotenuse }})= \\frac{\text{BC}}{\text{AC}})=\\frac{\text{1}}{\text{2}})
Let BC = k. Then, AC = 2k

Step 1: Compute Measure of side AB

Using Pythagoras theorem AB = \\sqrt{AC^2− BC^2 })
AB = \\sqrt{(2k)^2− k^2}) = \\sqrt{4k^2− k^2}) = \\sqrt{(3k)^2}) = \\sqrt{3k})

Step 2: Calculate the necessary trigonometric values

cos A = \\frac{\text{side adjacent to ∠A}}{\text{hypotenuse}}) =\\frac{\text{AB}}{\text{AC}}) =\\frac{\text{√3k}}{\text{2k}}) = \\frac{\sqrt{3}}{2})
sin C = \\frac{\text{side opposite to ∠C}}{\text{hypotenuse}})=\\frac{\text{AB}}{\text{AC}}) = \\frac{\text{√3k}}{\text{2k}}) = \\frac{\sqrt{3}}{2})
cos C =\\frac{\text{side adjacent to ∠C}}{\text{hypotenuse}}) =\\frac{\text{BC}}{\text{AC}})= \\frac{\text{k}}{\text{2k}}) = \\frac{\text{1}}{\text{2}})

Step 3: Compute value of given expressions

(i) sin C cos A – cos C sin A

[\\frac{\sqrt{3}}{2}) × \\frac{\sqrt{3}}{2})] - [\\frac{\text{1}}{\text{2}}) × \\frac{\text{1}}{\text{2}})] = [\\frac{\text{3}}{\text{4}}) - \\frac{\text{1}}{\text{4}})] = \\frac{\text{1}}{\text{2}})

(ii) cos A cos C + sin A sin C

[\\frac{\sqrt{3}}{2}) × \\frac{\text{1}}{\text{2}})] + [\\frac{\text{1}}{\text{2}}) × \\frac{\sqrt{3}}{2})] = \\frac{\sqrt{3}}{2})

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