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2 years ago

Sin10°-cos10° product and evaluate

Mathematics

2 answers:

Elena-2011 [213]2 years ago

8 0

$\\ \sf\longmapsto sin10°-cos10°$

$\\ \sf\longmapsto sin(90-10°)-cos10°$

sin(90-A)=cosA

$\\ \sf\longmapsto cos10-cos10$

$\\ \sf\longmapsto 0$

Montano1993 [528]2 years ago

6 0

Question should be sin 20 cos 10 + cos 20 sin 10

remember sin (a+b) = sin a cos b + cos a sin b ?

sin 20 cos 10 + cos 20 sin 10

= sin (20+10)

= sin 30

= 1/2

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It would be 4.08, so Ani is wrong writing it as 4.8.

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So, with this, the number 4 is 4, a single-digit number. Anything below that has a decimal point before it. With the knowledge of there being 8 hundredths, first see this diagram explaining the terms for numbers after the decimal point:

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Step-by-step explanaton:

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See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

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18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

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Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

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Sin10°-cos10° product and evaluate​

Sin10°-cos10° product and evaluate