Class 12 Inverse Trigonometric Functions · Case Study4 Markscase_based_application
Class 12 Maths · Inverse Trigonometric Functions · Case Study #34
For a school construction project, a triangle ABC has angles B=tan⁻¹(1/2) and C=tan⁻¹(1/3). Both angles are acute; the triangle is drawn in a separate SVG.
Part (1) 1 Marks
Find sin B.
Answer: 1/√5.
- Since tan B=1/2 and B is acute, use opposite 1, adjacent 2 and hypotenuse √5.
- Hence sin B=1/√5.
Part (2) 1 Marks
Find cos C.
Answer: 3/√10.
- Since tan C=1/3 and C is acute, use opposite 1, adjacent 3 and hypotenuse √10.
- Hence cos C=3/√10.
Part (3) 2 Marks
Find B+C.
Answer: π/4.
- Apply tan(B+C)=(tan B+tan C)/(1−tan B tan C).
- Substitution gives (1/2+1/3)/(1−1/6)=(5/6)/(5/6)=1. Since B,C are acute, B+C∈(0,π); positive tangent locates their sum in (0,π/2).
- Therefore B+C=tan⁻¹1=π/4.
Part (4) 2 Marks
Alternatively, find cos(B+C).
Answer: 1/√2.
- The acute-angle ratios give cos B=2/√5, sin B=1/√5, cos C=3/√10, sin C=1/√10.
- Use cos(B+C)=cos B cos C−sin B sin C=(6−1)/√50.
- Thus cos(B+C)=5/√50=1/√2.