Class 12 Inverse Trigonometric Functions · Case Study4 Markscase_based_application
Class 12 Maths · Inverse Trigonometric Functions · Case Study #33
A temple has top B and foot D, with BD=30 m. Two people stand at A and C on opposite sides along the same horizontal line A-D-C. The total separation AC=40√3 m and AD=30√3 m. Let α=∠CAB and β=∠BCA. Angles are measured from the horizontal ground; observer eye height is neglected.
Part (1) 1 Marks
Express α in terms of sin⁻¹.
Answer: α=sin⁻¹(1/2)=π/6.
- In right triangle ABD, AB=√((30√3)²+30²)=60 m, so sin α=BD/AB=1/2.
- As α is acute, α=sin⁻¹(1/2)=π/6.
Part (2) 1 Marks
Express α in terms of cos⁻¹.
Answer: α=cos⁻¹(√3/2)=π/6.
- AB=60 m, hence cos α=AD/AB=30√3/60=√3/2.
- Therefore α=cos⁻¹(√3/2)=π/6.
Part (3) 2 Marks
Express β in terms of tan⁻¹.
Answer: β=tan⁻¹√3=π/3.
- Since A-D-C are collinear, DC=AC−AD=(40−30)√3=10√3 m.
- In right triangle BDC, tan β=BD/DC=30/(10√3)=√3.
- As β is acute, β=tan⁻¹√3=π/3.
Part (4) 2 Marks
Alternatively, state the real domain and principal range of cos⁻¹x.
Answer: Domain [−1,1]; range [0,π].
- Cosine restricted to [0,π] is one-one and has image [−1,1].
- Its inverse therefore accepts inputs in [−1,1] and returns the unique angle in [0,π].
- Thus the arccosine domain is [−1,1] and its principal range is [0,π].