Question

In triangle ABC, if \( A = 45^\circ \), \( C = 75^\circ \), and \( R = \sqrt{2} \), then the value of \( r \) is:

• A. \( \frac{3 + \sqrt{3}}{\sqrt{3} + \sqrt{2} + 1} \)
• B. \( \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2} + 1} \)
• C. \( \frac{\sqrt{3}}{\sqrt{6} + \sqrt{3} + 3} \)
• D. \( \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2} + 1} \) \bigskip

Hint:

To determine the inradius, use the formula incorporating the sine of half the angles of the triangle. Ensure that trigonometric values for angles such as 22.5° and 37.5° are correctly applied for simplification.

Answer 1

The Correct Option is B

We are given a triangle \( ABC \) with: \[ A = 45^\circ, \quad C = 75^\circ, \quad R = \sqrt{2} \] We need to determine the inradius \( r \). --- Step 1: Compute \( B \) Using the angle sum property of a triangle: \[ B = 180^\circ - (A + C) \] \[ B = 180^\circ - (45^\circ + 75^\circ) = 60^\circ \] --- Step 2: Use the Formula for the Circumradius \( R \) The circumradius formula for a triangle is: \[ R = \frac{a}{2\sin A} \] Since we are not given side lengths explicitly, we proceed with the standard sine rule. From the sine rule: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \] Substituting \( R = \sqrt{2} \): \[ \frac{a}{\sin 45^\circ} = \frac{b}{\sin 60^\circ} = \frac{c}{\sin 75^\circ} = 2\sqrt{2} \] Evaluating sines: \[ \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \] Computing side lengths: \[ a = 2\sqrt{2} \times \frac{\sqrt{2}}{2} = 2 \] \[ b = 2\sqrt{2} \times \frac{\sqrt{3}}{2} = \sqrt{6} \] \[ c = 2\sqrt{2} \times \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{12} + \sqrt{4}}{2} = \frac{2\sqrt{3} + 2}{2} = \sqrt{3} + 1 \] --- Step 3: Compute Semi-Perimeter \( s \) \[ s = \frac{a + b + c}{2} \] \[ = \frac{2 + \sqrt{6} + \sqrt{3} + 1}{2} \] \[ = \frac{3 + \sqrt{6} + \sqrt{3}}{2} \] --- Step 4: Compute Inradius \( r \) The inradius formula: \[ r = \frac{\text{Area}}{s} \] Using Heron's formula: \[ \Delta = \frac{1}{2} ab \sin C \] \[ = \frac{1}{2} (2)(\sqrt{6}) \times \frac{\sqrt{6} + \sqrt{2}}{4} \] \[ = \frac{2\sqrt{6} (\sqrt{6} + \sqrt{2})}{8} \] \[ = \frac{12 + 2\sqrt{12}}{8} \] \[ = \frac{12 + 4\sqrt{3}}{8} = \frac{3 + \sqrt{3}}{2} \] Thus, \[ r = \frac{\Delta}{s} = \frac{\frac{3 + \sqrt{3}}{2}}{\frac{3 + \sqrt{6} + \sqrt{3}}{2}} \] Canceling \( \frac{1}{2} \): \[ r = \frac{3 + \sqrt{3}}{3 + \sqrt{6} + \sqrt{3}} \] This simplifies to: \[ r = \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2} + 1} \] --- Final Answer: \(\boxed{\frac{\sqrt{3}}{\sqrt{3} + \sqrt{2} + 1}}\)