Trigonometry! #87

Geometry Level 4

There exists a triangle satisfying the conditions

  1. b sin ⁡ A = a , A < π 2 b\sin A = a, A<\frac{\pi}{2}

  2. b sin ⁡ A > a , A > π 2 b\sin A > a, A>\frac{\pi}{2}

  3. b sin ⁡ A > a , A < π 2 b\sin A > a, A<\frac{\pi}{2}

  4. b sin ⁡ A < a , A < π 2 , b > a b\sin A < a, A<\frac{\pi}{2}, b>a

  5. b sin ⁡ A < a , A > π 2 , b = a b\sin A < a, A>\frac{\pi}{2}, b=a

Note: More than one option is correct. Enter the sum of the serial numbers of the correct options.

This problem is part of the set Trigonometry .


The answer is 5.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Shabarish Ch
Apr 25, 2015

Using sine rule, s i n A a = s i n B b \frac{sin A}{a} = \frac{sin B}{b}

b s i n A = a s i n B b sinA = a sinB

Now, a s i n B ≤ a a sinB \leq a . Equality holds when B = π 2 B = \frac{\pi}{2}

Therefore, 1 is possible. 2 and 3 are impossible. 4 is possible. 5 seems possible, but careful inspection reveals that 5 implies s i n A = s i n B sin A = sinB , which further implies A = B A=B or A = π − B A = \pi - B , both of which are impossible.

Hence the number to be input is 1 + 4 = 5 1 + 4 = \boxed{5}

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