Question

In a triangle ABC, if a = 4, b = 5 and c = 7, then \( \sin(A/2) = \)

• A. \( \frac{\sqrt{3}}{\sqrt{35}} \)
• B. \( \frac{\sqrt{35}}{3} \)
• C. \( \frac{\sqrt{2}}{\sqrt{35}} \)
• D. \( \frac{1}{\sqrt{35}} \)

Hint:

Half-angle formulas for sine in a triangle: \(\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}}\) \(\sin(B/2) = \sqrt{\frac{(s-a)(s-c)}{ac}}\) \(\sin(C/2) = \sqrt{\frac{(s-a)(s-b)}{ab}}\)
Where \(s = (a+b+c)/2\) is the semi-perimeter.

Answer 1

The Correct Option is D

We use the half-angle formula for \(\sin(A/2)\) in a triangle: \[ \sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}} \] where \(s\) is the semi-perimeter of the triangle, \(s = \frac{a+b+c}{2}\). Given \(a=4, b=5, c=7\). Calculate the semi-perimeter \(s\): \(s = \frac{4+5+7}{2} = \frac{16}{2} = 8\). Now calculate the terms needed: \(s-b = 8-5 = 3\). \(s-c = 8-7 = 1\). Substitute these into the formula: \[ \sin(A/2) = \sqrt{\frac{(3)(1)}{(5)(7)}} = \sqrt{\frac{3}{35}} = \frac{\sqrt{3}}{\sqrt{35}} \] This matches option (a). Let me recheck the options and provided answer. The checkmark in the image is on option (a) \(\frac{\sqrt{3}}{\sqrt{35}}\). My calculation yields \(\frac{\sqrt{3}}{\sqrt{35}}\). Is it possible the question has a typo or my formula application for the specific marked option is wrong? Let's check if option (d) \(1/\sqrt{35}\) can be obtained. If \(\sin(A/2) = 1/\sqrt{35}\), then \((s-b)(s-c)/(bc) = 1/35\). \((s-b)(s-c) = bc/35 = (5)(7)/35 = 35/35 = 1\). With \(s=8\), \(s-b = 8-5 = 3\), \(s-c = 8-7 = 1\). So \((s-b)(s-c) = 3 \times 1 = 3\). This means \(3 = 1\), which is false. So option (d) is incorrect based on my calculation. Option (a) is correct. It seems the provided solution might be (a) according to my calculation. The checkmark in the image itself for question 26 is quite faint but appears to be on option (a) (the topmost option). If the correct answer (as per an external key if different from image) was indeed (d) \(1/\sqrt{35}\), then the problem statement values (a,b,c) would need to be different for \((s-b)(s-c)=1\). For example, if \(s-b=1\) and \(s-c=1\). \(8-b=1 \implies b=7\). \(8-c=1 \implies c=7\). If \(b=7, c=7, a=2\) (to make \(s=8\)). Then \(a=2, b=7, c=7\). \(\sin(A/2) = \sqrt{\frac{(8-7)(8-7)}{7 \times 7}} = \sqrt{\frac{1 \times 1}{49}} = \frac{1}{7}\). Not \(1/\sqrt{35}\). My calculation leads to \(\sin(A/2) = \sqrt{3/35}\). I will use this. The image's checkmark for Q26 is on the first option which is \(\frac{\sqrt{3}}{\sqrt{35}}\). My solution matches this. \[ \boxed{\frac{\sqrt{3}}{\sqrt{35}}} \]