Question

Solve for a side in right triangles.

Answer 1

Answer:   " 1. 71 " .  The length of side BC is:  " 1. 71 units " .

The length of side BC is:  "  .The length of side BC is:  " 1. 71 units" " .

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Step-by-step explanation:

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Note that for "right triangles" ; that is, triangles that have one right angle (one angle with a measurement of 90° ) ; such as  in the "image given" —the right angle is ∠BCA — we can use the mnemonic :

        SOH -  CAH - TOA  ;

→  which stands for the following:

The "sin" of angle = {opposite side length] / [hypotenuse length] ;  

               that is:  "sin = opp/hyp" ;

The "cos" of an angle = {adjacent side length] / [hypotenuse length] ;

               that is:  "cos = adj/ hyp"  ;

The "tan"  of an angle = {opposite side length] / {adjacent side length] ;

               that is:   "tan = opp/adj" .

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In the image attached:

We are given the measurement of another angle—  70° .

We are also given the hypotenuse length, which is: " 5 " [units].

We as to solve for the unknown side length; that is, the side length of BC .

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Let us start by looking at the given measurement of the particular angle,    

  70° —which is:  m∠B    The unknown side length, BC, is adjacent to ∠B .

We are given (from the "image attached")  the hypotenuse side length; which is:   " 5 " .

Now, using the mnemonic device:  SOH CAH TOA ; can we solve for the unknown side length BC ; which is "Adjacent" to ∠B ?  

Consider the mnemonic:  SOH CAH TOA .

The "S" stands for "sin" ; the "C" stands for "cos" ;

the "T" stands for "tangent" ;

The "sin" —SOH—sin = opp/hyp" —does not include the "adjacent" , whereas there other 2 (two):  CAH ; and TOA —do include the adjacent.

So we can rule out using the "sin" to solve our problem.

Now, continue with the "cos"—CAH—cos = adj/ hyp " ;

 →  cos  = (adjacent side length / hypotenuse side length) ;

This portion of the mnemonic includes the "adj" ["adjacent side length"] — [i.e. the length of side BC—which is the value for which we which are to solve!)  

Furthermore, this part of the "mnemonic" includes the "hypotenuse side length" ;

→  CAH ;   →  cos = adj / hyp. ;  as aformentioned ;  

→ so we can plug in our given value—" 5"— for the "hypotenuse" !  [Note that this "given value" is shown within the "attached image"—directly along the "hypotenuse" side of the right triangle shown this "attached image".].

As such, we shall solve the problem; that is; obtain the side length of BC ;

using the:  "CAH" ;  → cos = adj / hyp ;  {portion of the mnemonic}; as aforementioned:

→  cos ∠B  = cos 70 ;  = [adjacent side light] / [hypotenuse side length] ;

=  adj / hyp .     →  {Added note:  abbreviations.}.

→ cos 70  = [unknown side length of BC} / 5  ;

Now, solve for the "[unknown side length of BC]" ;  which is the answer to this very "Brainly" question being asked.

Note:  For the sake of convenience, let us represent—and write— this value (for we shall solve); as:  " x " .

→ cos 70  = x / 5  ;   →  Solve for "x" .   ↔  x / 5  =  cos 70 ;  Solve for "x" .

Multiply each side of the equation by:  "5" ;  as follows:

5* (x/5)  = 5 * cos 70 ; to get:

→  x = 5 cos 70 = 5 * (cos 70) =  5 * (0.342020143326) ;

= 1.71010071663 ;  Now:  round down: to 2 (two) decimal places:

=  1.71  units ;  which is our answer.

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         If that number —i.e. "in the thousandths place"—has a digit that is 0, 1, 2, 3, or 4; then we round "down"; and the value—when rounded to the nearest hundredths —is written with to the precision that includes the nearest two (2) decimals only (in terms of number of decimal spaces retained—& in manner that includes the same "two (2) decimal digits occurs in the original value.  {Exception:  If there are less than 2 (two) decimal digits left, one or two decimal digits consisting of the number—"[0}"—may be added (when and if necessary) —to allow for exactly "2" (two) decimal digits with the digit—"0" [zero]—no more/no less—when rounding to the nearest hundredth.}.

        If that number —i.e. "in the thousandths place"—has a digit that is 5, 6, 7, 8, 9 ; then we round "up"; and the decimal value in the "hundredths" placed is rounded up—& the value in the nearest hundredths place is changed to the next digit.

If the value in the "thousandths place" at least 5 but no more than 9 ; AND: the value of the "hundredths place" is "9" ; then the value is rounded up in a manner in which both the "tenths" place AND the "hundredths" place changed.o, our answer for the problem:

  →   " 1.71010071663 " ;  Now: Round to 2 (two) decimal places:

Consider: " 1.710.."

→ The "thousandths place" has a digit of "zero" ; and the hundredth place has a digit of:  "1" ; which less than "9" ;

→  So we can round our value down;  to the nearest hundredths, to:  " 1.71 " .

The answer is:  1.71 units.