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Question:

In a ΔABC, ab = 2 + √3 and ∠C = 60°. Then the ordered pair (∠A, ∠B) is equal to: (45°, 75°), (75°, 45°), (105°, 15°), (15°, 105°)

(105o,15o)

(75o,45o)

(45o,75o)

(15o,105o)

Solution:

We have A + B + C = 180° ⇒ A + B = 120° (1)
Using sine rule in ΔABC,
\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}
\frac{a}{sinA} = \frac{b}{sinB} = \frac{ab}{sinA sinB} = \frac{2 + \sqrt{3}}{sinA sinB}
Also, \frac{ab}{sinC} = \frac{ab}{sin60°} = \frac{2 + \sqrt{3}}{\frac{\sqrt{3}}{2}} = \frac{2(2 + \sqrt{3})}{\sqrt{3}} = \frac{4 + 2\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3} + 6}{3} = \frac{2(2\sqrt{3} + 3)}{3}
From sine rule,
\frac{c}{sin60°} = \frac{a}{sinA} = \frac{b}{sinB}
We have a = c \frac{sinA}{sin60°} and b = c \frac{sinB}{sin60°}
a b = c^2 \frac{sinA sinB}{sin^2 60°} = 2 + \sqrt{3}
Also, A + B = 120°
If A = 75°, B = 45°, then sinA sinB = sin75° sin45° = \frac{\sqrt{3} + 1}{2\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{4}
If A = 45°, B = 75°, then sinA sinB = sin45° sin75° = \frac{1}{\sqrt{2}} \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{4}
If A = 105°, B = 15°, then sinA sinB = sin105° sin15° = sin(60° + 45°) sin(45° - 30°) = (\frac{\sqrt{3} + 1}{2\sqrt{2}}) (\frac{\sqrt{3} - 1}{2\sqrt{2}}) = \frac{3 - 1}{8} = \frac{1}{4}
If A = 15°, B = 105°, then sinA sinB = sin15° sin105° = \frac{1}{4}
Since ab = 2 + √3, then sinA sinB = 1/4 is not possible. Hence A = 75°, B = 45° or A = 45°, B = 75°