A question on trigonometry

Geometry Level pending

If 0 ∘ < θ < 9 0 ∘ 0^\circ < \theta < 90^\circ and csc ⁡ θ = 13 12 \csc \theta = \dfrac{13}{12} , find 2 sin ⁡ θ − 3 cos ⁡ θ 4 sin ⁡ θ − 9 cos ⁡ θ \dfrac{2\sin \theta - 3\cos \theta}{4\sin \theta - 9\cos \theta} .


The answer is 3.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
Dec 15, 2020

Given that csc ⁡ θ = 13 12 = 1 sin ⁡ θ ⟹ sin ⁡ θ = 12 13 \csc \theta = \dfrac {13}{12} = \dfrac 1{\sin \theta} \implies \sin \theta = \dfrac {12}{13} . Since cos ⁡ θ = 1 − sin ⁡ 2 θ = 1 − 1 2 2 1 3 2 = 169 + 144 169 = 25 169 = 5 13 \cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1-\dfrac {12^2}{13^2}} = \sqrt{\dfrac {169+144}{169}} = \sqrt{\dfrac {25}{169}} = \dfrac 5{13} . Then we have:

2 sin ⁡ θ − 3 cos ⁡ θ 4 sin ⁡ θ − 9 cos ⁡ θ = 2 ( 12 ) − 3 ( 5 ) 4 ( 12 ) − 9 ( 5 ) = 24 − 15 48 − 45 = 9 3 = 3 \frac {2\sin \theta - 3\cos \theta}{4\sin \theta - 9\cos \theta} = \frac {2(12)-3(5)}{4(12)-9(5)} = \frac {24-15}{48-45} = \frac 93 = \boxed 3

4 Helpful 1 Interesting 2 Brilliant 1 Confused

We know that sinθ = o p p o s i t e s i d e h y p o t e n u s e \frac{opposite side}{hypotenuse} and cosθ = a d j a c e n t s i d e h y p o t e n u s e \frac{adjacent side}{hypotenuse} . From this diagram,

sinθ = 12 13 \frac{12}{13} cosθ = 5 13 \frac{5}{13}

From this, 2. s i n θ − 3. c o s θ 4. s i n θ − 9. c o s θ \frac{2.sinθ - 3.cosθ }{4.sinθ - 9.cosθ} can be simplified as 3 \boxed{3}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nitpick: you have not constrained 0 < θ < π 2 0 < \theta < \frac{\pi}{2} . It could be the case that cos ⁡ ( θ ) = − 5 13 \cos(\theta) = \frac{-5}{13}

Richard Desper - 5 months, 4 weeks ago

Sorry, I forgot that.

A Former Brilliant Member - 5 months, 4 weeks ago

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