Triangle ABC has AB = 10 cm, BC = 7 cm, and angle ABC = \(\theta\) radians. The area of triangle ABC is 25 cm².
(a) Find the value of \(\sin\theta\). [2]
(b) Given that \(\theta\) is obtuse, find the value of \(\cos\theta\). [3]
(c) Find the length of AC. [3]
Point D lies on AC such that BD bisects angle ABC.
(d) Find the size of angle ABD. [2]
(e) Using the sine rule in triangle ABD, find the length of BD. [3]
Question
Find the value of \(\sin\theta\). [2]
Hint
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Use the area formula \(A = \frac{1}{2} \times AB \times BC \times \sin\theta\).
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\(\sin\theta = \frac{5}{7}\)
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Given that \(\theta\) is obtuse, find the value of \(\cos\theta\). [3]
Hint
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Hint
Apply the identity \(\cos^2\theta = 1 – \sin^2\theta\) and note that \(\cos\theta\) is negative when \(\theta\) is obtuse.
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\(\cos\theta = -\frac{2\sqrt{6}}{7}\)
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Find the length of AC. [3]
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Apply the cosine rule: \(AC^2 = AB^2 + BC^2 – 2\times AB \times BC \times \cos\theta\).
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\(AC = \sqrt{149+40\sqrt6} \approx 15.7\;cm\)
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Find the size of angle ABD. [2]
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Recognize that BD bisects angle ABC, so \(\angle ABD = \frac{\theta}{2}\).
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\(\angle ABD = \frac{\theta}{2} \approx 1.17\;radians\)
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Using the sine rule in triangle ABD, find the length of BD. [3]
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Set up the sine rule in triangle ABD as \(\frac{BD}{\sin(\angle BAC)} = \frac{AB}{\sin(\angle ADB)}\) and substitute the known values.
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\(BD \approx 3.20\;cm\)
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