Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Trigonometric Ratios
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Exercise 47 Page 651

Practice makes perfect
a We can simplify the given expression using the definition of the cosine ratio and the tangent ratio. Let's start by recalling these ratios.
cos(θ) Adjacent leg/Hypotenuse Adj/Hyp
tan(θ) Opposite leg/Adjacent leg Opp/Adj
Now, we will use these ratios to simplify the given product. Let's do it!
cos A * tan A
( Adj/Hyp ) * ( Opp/Adj )
Adj* Opp/Hyp * Adj
Adj * Opp/Hyp * Adj
Opp/Hyp
Notice that the ratio that we found corresponds to the definition of the sine ratio. Opposite leg/Hypotenuse ⇔ sin A
b We will use the definition of the sine ratio and the tangent ratio to simplify the given expression.
sin(θ) Opposite leg/Hypotenuse Opp/Hyp
tan(θ) Opposite leg/Adjacent leg Opp/Adj
Let's do it!
sin A Ă· tan A
( Opp/Hyp ) Ă· ( Opp/Adj )
(Opp/Hyp) * (Adj/Opp)
Opp* Adj/Hyp * Opp
Opp * Adj/Hyp * Opp
Adj/Hyp
Notice that the ratio that we found corresponds to the definition of the cosine ratio. Adjacent leg/Hypotenuse ⇔ cos A
c One more time, we will use the definition of the sine ratio and the cosine ratio to simplify the given expression.
sin(θ) Opposite leg/Hypotenuse Opp/Hyp
cos(θ) Adjacent leg/Hypotenuse Adj/Hyp
Let's do it!
sin A Ă· cos A
( Opp/Hyp ) Ă· ( Adj/Hyp )
(Opp/Hyp) * (Hyp/Adj)
Opp* Hyp/Hyp * Adj
Opp * Hyp/Hyp * Adj
Opp/Adj
Notice that the ratio that we found corresponds to the definition of the tangent ratio. Opposite leg/Adjacent leg ⇔ tan A