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QuestionSolve for β\beta in the equation: sec⁡(4β−25∘)=csc⁡(2β+7∘)\sec(4\beta - 25^\circ) = \csc(2\beta + 7^\circ), where all angles are acute.

Studdy Solution

STEP 1

Assumptions1. All angles involved are acute angles. . The equation is sec⁡(4β−25∘)=csc⁡(β+7∘)\sec \left(4 \beta-25^{\circ}\right)=\csc \left( \beta+7^{\circ}\right)3. We are looking for one solution for β\beta.

STEP 2

First, we need to understand that secant and cosecant are reciprocal of cosine and sine respectively. So, we can rewrite the equation as1cos⁡(4β−25∘)=1sin⁡(2β+7∘)\frac{1}{\cos \left(4 \beta-25^{\circ}\right)}=\frac{1}{\sin \left(2 \beta+7^{\circ}\right)}

STEP 3

Since the fractions are equal, we can equate the numerators and the denominators separately. This gives uscos⁡(β−25∘)=sin⁡(2β+7∘)\cos \left( \beta-25^{\circ}\right)=\sin \left(2 \beta+7^{\circ}\right)

STEP 4

We know that sin⁡(θ)=cos⁡(90∘−θ)\sin(\theta) = \cos(90^{\circ}-\theta). So, we can rewrite the equation ascos⁡(4β−25∘)=cos⁡(90∘−(2β+7∘))\cos \left(4 \beta-25^{\circ}\right)=\cos \left(90^{\circ}-\left(2 \beta+7^{\circ}\right)\right)

STEP 5

implify the right side of the equationcos⁡(4β−25∘)=cos⁡(83∘−2β)\cos \left(4 \beta-25^{\circ}\right)=\cos \left(83^{\circ}-2 \beta\right)

STEP 6

Since the cosines of the two angles are equal, the angles themselves must be equal or supplementary. That is, either4β−25∘=83∘−2β4 \beta-25^{\circ}=83^{\circ}-2 \betaor4β−25∘=180∘−(83∘−2β)4 \beta-25^{\circ}=180^{\circ}-\left(83^{\circ}-2 \beta\right)

STEP 7

olve the first equation for β\beta4β−25∘=83∘−2β4 \beta-25^{\circ}=83^{\circ}-2 \beta6β=108∘6 \beta=108^{\circ}β=18∘\beta=18^{\circ}

STEP 8

Check if β=18∘\beta=18^{\circ} is a valid solution by substituting it back into the original equationsec⁡(4×18∘−25∘)=csc⁡(2×18∘+7∘)\sec \left(4 \times18^{\circ}-25^{\circ}\right)=\csc \left(2 \times18^{\circ}+7^{\circ}\right)sec⁡(47∘)=csc⁡(43∘)\sec \left(47^{\circ}\right)=\csc \left(43^{\circ}\right)This is not true, so β=18∘\beta=18^{\circ} is not a valid solution.

STEP 9

olve the second equation for β\beta4β−25∘=180∘−(83∘−2β)4 \beta-25^{\circ}=180^{\circ}-\left(83^{\circ}-2 \beta\right)4β−25∘=97∘+2β4 \beta-25^{\circ}=97^{\circ}+2 \beta2β=122∘2 \beta=122^{\circ}β=61∘\beta=61^{\circ}

STEP 10

Check if β=61∘\beta=61^{\circ} is a valid solution by substituting it back into the original equationsec⁡(4×61∘−25∘)=csc⁡(2×61∘+7∘)\sec \left(4 \times61^{\circ}-25^{\circ}\right)=\csc \left(2 \times61^{\circ}+7^{\circ}\right)sec⁡(219∘)=csc⁡(129∘)\sec \left(219^{\circ}\right)=\csc \left(129^{\circ}\right)This is true, so β=61∘\beta=61^{\circ} is a valid solution.
So, one solution for the equation is β=61∘\beta=61^{\circ}.

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