Trigno+geometry

Geometry Level 4

{ 3 sin ⁡ A + 4 cos ⁡ B = 6 3 cos ⁡ A + 4 sin ⁡ B = 1 \large \begin{cases}{3\sin A + 4\cos B = 6} \\ {3\cos A + 4\sin B = 1}\end{cases}

In a triangle A B C ABC , we are given that two angles A , B A,B satisfy the equation above, find the measure of the remaining angle, C C (in degrees).

90 150 60 30

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Avn Bha by avn bha , 22, India

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1 solution

Chew-Seong Cheong
May 24, 2015

{ 3 sin ⁡ A + 4 cos ⁡ B = 6 3 cos ⁡ A + 4 sin ⁡ B = 1 ⇒ { ( 3 sin ⁡ A + 4 cos ⁡ B ) 2 = 6 2 ( 3 cos ⁡ A + 4 sin ⁡ B ) 2 = 1 2 ⇒ { 9 sin ⁡ 2 A + 24 sin ⁡ A cos ⁡ B + 16 cos ⁡ 2 B = 36 9 cos ⁡ 2 A + 24 sin ⁡ B cos ⁡ A + 16 sin ⁡ 2 B = 1 ⇒ 9 + 24 ( sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A ) + 16 = 37 24 ( sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A ) = 12 ⇒ sin ⁡ ( A + B ) = 1 2 ⇒ A + B = 3 0 ∘ or 15 0 ∘ \begin{cases} 3\sin{A} +4\cos{B} = 6 \\ 3\cos{A} +4\sin{B} = 1 \end{cases} \\ \Rightarrow \begin{cases} (3\sin{A} +4\cos{B})^2 = 6^2 \\ (3\cos{A} +4\sin{B})^2 = 1^2 \end{cases} \\ \Rightarrow \begin{cases} 9\sin^2{A} +24\sin{A}\cos{B} + 16 \cos^2{B} = 36 \\ 9\cos^2{A} +24\sin{B}\cos{A} + 16\sin^2{B} = 1 \end{cases} \\ \Rightarrow 9 + 24(\sin{A}\cos{B} + \sin{B}\cos{A}) + 16 = 37 \\ \quad \space 24(\sin{A}\cos{B} + \sin{B}\cos{A}) = 12 \\ \Rightarrow \sin{(A+B)} = \frac{1}{2} \\ \Rightarrow A+B = 30^\circ \text{ or } 150^\circ

If A + B = 3 0 ∘ A+B = 30^\circ , from 3 cos ⁡ A + 4 sin ⁡ B = 1 3\cos{A} +4\sin{B} = 1 , we have:

sin ⁡ B = 1 − 3 cos ⁡ A 4 < 1 − 3 × 3 2 4 < 0 ⇒ B > 18 0 ∘ Rejected \sin{B} = \dfrac {1 - 3\cos{A}}{4} < \dfrac {1 - 3\times \frac{\sqrt{3}}{2}}{4} < 0 \quad \Rightarrow B > 180^\circ \text{ Rejected}

⇒ A + B = 15 0 ∘ ⇒ C = 18 0 ∘ − 15 0 ∘ = 3 0 ∘ \Rightarrow A+B = 150^\circ \quad \Rightarrow C = 180^\circ - 150^\circ = \boxed{30^\circ}

5 Helpful 0 Interesting 0 Brilliant 1 Confused

Moderator note:

Great job with rejecting A + B = 3 0 ∘ A + B = 30 ^ \circ .

In a similar way, could we use the condition of 3 sin ⁡ A + 4 cos ⁡ B = 6 3 \sin A + 4 \cos B = 6 to reject the 3 0 ∘ 30 ^ \circ case?

Challenge Master : If A+B =30 , then A will be less than 30 (in degrees). Thus, sinA will be less than 1/2. This would imply that cosB will be greater than 1 , which is not possible. Hence A+B =150 is correct while A+B =30 is contradictory.

Abhijeet Verma - 6 years ago

i didn't understand the last step that you did sir ? how and why did you reject 30

avn bha - 6 years ago

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Sorry, it should be B > 18 0 ∘ B > 180^\circ , because sin ⁡ B < 0 \sin{B} < 0 .

Chew-Seong Cheong - 6 years ago

Do you mean the condition of 3 cos ⁡ A + 4 sin ⁡ B = 6 3\cos{A} + 4\sin{B} = 6 ?

Chew-Seong Cheong - 6 years ago

if A+B= 30, ------>A= 30 - B

3 sin (30-B)+4 cos B =6,

After the calculation, we get,

5.5 cos B - (1.732/2)sin B = 6 -------> for this condition to be satisfied sin B should be a negative value and hence B should be greater than 180.......which is contradictory to the properties of Triangle.

Hence A + B = 30 is ruled out.

Anil Krishna - 5 years, 10 months ago

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