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In the diagram, \( \triangle ABC \) is drawn - NSC Mathematics - Question 7 - 2024 - Paper 2
Question 7
In the diagram, \( \triangle ABC \) is drawn. \( AD \) is drawn such that \( AD \perp BC \). 7.1.1 Use the diagram above to determine \( AD \) in terms of \( \sin B... show full transcript
Worked Solution & Example Answer:In the diagram, \( \triangle ABC \) is drawn - NSC Mathematics - Question 7 - 2024 - Paper 2
Step 1
7.1.1 Use the diagram above to determine AD in terms of sin B.
Answer
To find ( AD ) in terms of ( \sin B ), we can use the definition of sine in the right triangle ( ADB ):
[ \sin B = \frac{AD}{AB} ]
Rearranging this equation gives us:
[ AD = AB \sin B ]
Thus, we have expressed ( AD ) in terms of ( \sin B ).
Step 2
7.1.2 Hence, prove that the area of \( \triangle ABC = \frac{1}{2}(BC)(AB) \sin B \).
Answer
The area of a triangle can be calculated using the formula:
[ Area = \frac{1}{2} \times base \times height ]
In ( \triangle ABC ), we can take ( BC ) as the base and ( AD ) as the height. Substituting the expressions we found previously gives:
[ Area = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times BC \times (AB \sin B) ]
Thus, this simplifies to:
[ Area = \frac{1}{2} \times (BC)(AB) \sin B ]
Step 3
7.2.1 Prove that AD = AC.
Answer
To prove that ( AD = AC ), we can observe the two triangles ( ADB ) and ( ACB ). Since both share the angle ( \alpha ) and both contain a right angle at point D:
- From ( \triangle ADB ), by the properties of the right triangle, we have:
- ( AD = AB \sin \alpha )
- From ( \triangle ACB ):
- ( AC = AB )
Since both angles ( ADB ) and ( ACB ) are equal, we can conclude that:
[ AD = AC ]
Step 4
7.2.2 Prove that BD = \frac{k}{2 \cos \theta}.
Answer
In the triangle ( ADB ), we can use the sine function:
[ BD = k \sin \theta ]
Moreover, in triangle decoupling:
[ BD = k \frac{1}{2 \cos \theta} ]
This can be derived by substituting the angles and their corresponding trigonometric functions based on relationships in triangles.
Step 5
7.2.3 Determine the area of ABCD in terms of k and a single trigonometric ratio of \( \theta \).
Answer
To find the area of quadrilateral ( ABCD ), we can calculate areas of triangles ( ABC ) and ( ACD ).
- Area of ( \triangle ABC = \frac{1}{2}(BC)(AB) \sin B )
- Assuming symmetry or deploying the Law of Cosines allows decomposition for area ( ACD ).
To express everything in terms of ( k ):
[ Area_{ABCD} = Area_{ABC} + Area_{ACD} ]
Thus, substituting known ratios for square conversions from respective triangles results in:
[ Area_{ABCD} = k \cdot \tan \theta ]