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Question # (1):   t = 15.8 (when rounded to the nearest tenth).
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Question # (3):  The length of Side EF is 7.5 (when rounded to the nearest tenth), and the length of side DF is 2.7 (when rounded to the nearest tenth).
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Here is how to solve for “Question #1” and “Question #3”:
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Remember the phrase: “SOH CAH TOA”; in which:
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Sin = opposite/hypotenuse; Cos = adjacent/ hypotenuse;
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We are given an triangle; one angle has a measurement of 44°.  Considering this given angle, the respective hypotenuse given is “22”—along with a variable “t” for the respective “adjacent” side.

Using the above: “SOH CAH TOA” method; it would make sense to use the “CAH” {cosine = adjacent/hypotenuse} part of the mnemonic to solve for “t”.
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Reason: We are given the angle measure 44
°; the hypotenuse value (“22”); so the cosine of 44, or “cos 44” = adjacent/hypotenuse = “t” / 22 .
→ Write as:  cos 44 = t / 22 ;  Solve for “t” .
→ (cos 44)*22 = t ; → t = 22 * (cos 44);
→  t = 22 *(cos 44) = 22 * (0.719339800339) → Use a scientific calculator:
→  t = 15.825475607458 ;  → Round to the nearest tenth, as indicated in the problem, to get:
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→ t = 15.8 ; → which is the answer.
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“Question #3: Find the missing lengths in the triangle. Round to the nearest tenth.” [Note: There are 2 (two) missing lengths to find.].
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To get the length of side “EF”; let us use the “CAH” part of the mnemonic,
“SOH CAH TOA”, (in the aforementioned previous problem given).

The “CAH” stands for the formula/property to calculate for the “cosine”—specifically, “cosine = adjacent/hypotenuse”.  We choose this one because we are given the measure of “angle E = 20

° “; and we are given the hypotenuse, “side ED = 8”; and we want to solve for the adjacent side, which is “EF” (which we will call: “x”).
→ So, cos 20 = adjacent/ hypotenuse = x /8; → Solve for “x””
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(cos 20)*8 = x ;
→  8*(cos 20) = x ; →
→x = 8 * (cos 20) = 8 *(0.939692620786) → Use a scientific calculator:                 = 7.517540966288 ;→ Round to the nearest tenth, as indicated in the problem, to get:
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→ 7.5 ; which is the answer for the length of side EF.
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Now, we need to find the unknown length of the other side, “Side DF”.
Here is one method:
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Find the measure of “Angle D”.  Note that for any triangle, all three sides must add up to 180
°.  We know that this is a right triangle, with one angle (“Angle F”) having a measurement of 90°.  We are also given that another angle (“Angle E”) has a measurement of 20°.
→ So, the measurement of “Angle D” = 180 – (90 + 20) = 180 – 110 = 70.
→ We want to solve for the length of side “DF”, which is adjacent to “Angle D” (which we know is 70°).
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Let us use
the “CAH” part of the mnemonic,
“SOH CAH TOA”, (in the aforementioned previous problem given).
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The “CAH” stands for the formula/property to calculate for the “cosine”—specifically, “cosine = adjacent/hypotenuse”.  We choose this one because we are given the measure of angle “D”= 70°; and we are given the hypotenuse = 8. The “adjacent”, or “Side DF”, is the unknown value (let us make it “x”).
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So, cos 70 = adjacent / hypotenuse = x/ 8 ; Solve for “x” :
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cos 70 = x / 8;  (cos 70) * 8 = x ;   8 * (cos 70) = x;
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x = 8 * (cos 70) = 8 * (0.342020143326) → Use a scientific calculator →             = 2.736161146608 ;
→Round to the nearest tenth, as indicated in the problem, to get:
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→ 2.7 ; which is our answer, for the length of side DF.
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So, the answer for “Question 1” is: t = 15.8 (when rounded to the nearest tenth).
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For “Question #3”, the length of Side EF is 7.5 (when rounded to the nearest tenth), and the length of side DF is 2.7 (when rounded to the nearest tenth).