Math Problem Statement
Given, Triangle ABC has angle C = θ, AC=1, BC=x (i) If {(AC/BC+ AB/BC)÷(AC/BC- AB/BC)} = (AC+AB)/BC, prove that, AB/AC = √(AC-BC)/√(AC+BC)
Solution
Given the equation:
we need to prove that:
Step 1: Simplify the left-hand side (LHS) of the given equation
We begin by simplifying the LHS of the equation:
Since both the numerator and the denominator contain terms with and , we can factor out :
Step 2: Simplify the right-hand side (RHS)
The RHS of the given equation is:
Step 3: Set LHS equal to RHS
From step 1, the LHS becomes:
Equating this to the RHS from step 2:
This implies that:
So we have the key relationship:
Step 4: Substitute into the required proof
Now, using this result, we substitute into the expression we want to prove, i.e., .
Since , the left-hand side (LHS) becomes:
Now, check if this equals the RHS, which is . Squaring both sides:
Clearly, this holds true, so we have:
Thus, the equation is proven.
Conclusion:
Would you like a more detailed explanation of any step? Here are some related questions for further exploration:
- How does this equation relate to the properties of triangles?
- Can this method of solving be applied to other geometric problems?
- What role does the Pythagorean theorem play in such triangle-based proofs?
- How does trigonometry come into play in such proofs involving angles and sides?
- How would the proof change if and were given numerical values?
Tip: In triangle proofs, always identify key relationships between angles and sides early to simplify the algebra.
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Geometry
Ratios and Proportions
Formulas
Ratio identity: (AC/BC + AB/BC) ÷ (AC/BC - AB/BC) = (AC + AB)/BC
Proportion involving triangle sides: AB/AC = √(AC - BC)/√(AC + BC)
Theorems
Triangle Proportionality Theorem
Algebraic manipulation of proportions
Suitable Grade Level
Grades 10-12
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