Solve for a side in right triangles.

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.