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Question
Math
Posted 7 months ago

Find cos⁡(∠ABC−60∘)\cos \left(\angle A B C-60^{\circ}\right).
Choose 1 answer:
(A) 310+253\frac{3}{10}+\frac{2}{5} \sqrt{3}
(B) 25−3103\frac{2}{5}-\frac{3}{10} \sqrt{3}
(C) 25+3103\frac{2}{5}+\frac{3}{10} \sqrt{3}
() 310−253\frac{3}{10}-\frac{2}{5} \sqrt{3}
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 2
Given that ∠ABC \angle ABC is part of a right triangle, we can find cos⁡∠ABC \cos \angle ABC by using the adjacent side over the hypotenuse: cos⁡∠ABC=BCAB=810=45 \cos \angle ABC = \frac{BC}{AB} = \frac{8}{10} = \frac{4}{5}
step 3
We also know that cos⁡60∘=12 \cos 60^{\circ} = \frac{1}{2} and sin⁡60∘=32 \sin 60^{\circ} = \frac{\sqrt{3}}{2}
step 4
Applying the cosine subtraction formula: cos⁡(∠ABC−60∘)=cos⁡∠ABC⋅cos⁡60∘+sin⁡∠ABC⋅sin⁡60∘ \cos(\angle ABC - 60^{\circ}) = \cos \angle ABC \cdot \cos 60^{\circ} + \sin \angle ABC \cdot \sin 60^{\circ}
step 5
Substituting the known values: cos⁡(∠ABC−60∘)=45⋅12+sin⁡∠ABC⋅32 \cos(\angle ABC - 60^{\circ}) = \frac{4}{5} \cdot \frac{1}{2} + \sin \angle ABC \cdot \frac{\sqrt{3}}{2}
step 6
To find sin⁡∠ABC \sin \angle ABC , we use the opposite side over the hypotenuse: sin⁡∠ABC=ACAB=610=35 \sin \angle ABC = \frac{AC}{AB} = \frac{6}{10} = \frac{3}{5}
step 7
Now substitute sin⁡∠ABC \sin \angle ABC into the equation: cos⁡(∠ABC−60∘)=45⋅12+35⋅32 \cos(\angle ABC - 60^{\circ}) = \frac{4}{5} \cdot \frac{1}{2} + \frac{3}{5} \cdot \frac{\sqrt{3}}{2}
step 8
Simplify the expression: cos⁡(∠ABC−60∘)=25+35⋅32 \cos(\angle ABC - 60^{\circ}) = \frac{2}{5} + \frac{3}{5} \cdot \frac{\sqrt{3}}{2}
step 9
Combine the terms: cos⁡(∠ABC−60∘)=25+3310 \cos(\angle ABC - 60^{\circ}) = \frac{2}{5} + \frac{3\sqrt{3}}{10}
1 Answer
C
Key Concept
Cosine subtraction formula
Explanation
The cosine subtraction formula allows us to find the cosine of the difference between two angles. In this case, it is used to find cos⁡(∠ABC−60∘) \cos(\angle ABC - 60^{\circ}) by knowing the cosine and sine of ∠ABC \angle ABC and the cosine and sine of 60∘ 60^{\circ} .

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