The answer is: " [tex]m[/tex]∠[tex]TPQ[/tex] = 124° " .
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Step-by-step explanation:
We are asked to find: "m" (the "measurement");
specifically, " [tex]m[/tex]∠[tex]TPQ[/tex] " ;
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→ which, from the diagram given:
is represented by "(11x + 14)" .
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→ If we can solve for the value of "x" ; then we can solve for:
"(11x + 14)" ; i.e. " [tex]m[/tex]∠[tex]TPQ[/tex] " .
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Let us consider " ∡[tex]QPR[/tex] " ;
→ which is the "supplementary angle" to: " ∡[tex]TPQ[/tex] " :
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That is: " [tex]m[/tex]∠[tex]TPQ[/tex] + [tex]m[/tex]∠[tex]QPR[/tex] = 180° " .
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Note: By definition, "supplementary angles" add up to: "180° " ;
→ even if multiple angles are involved.
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Note: "∡[tex]TPQ[/tex] " and " ∡[tex]QPR[/tex] " —
together — form a "straight line" ;
→ which means that the 2 (two) angles are
"supplementary" ; and:
→ which means that the: sum of the measurements of
the " 2 (two) angles " —equal: " 180° .
{Note: by "forming a straight line" ; for this purpose, this criterion also is satisfied by forming a "straight line on a line segment" —even if that "line segment" is actually:
1) an actual "line segment" ; or:
2) a portion of a "line segment" ;
3) a "line segment" that is actual part of a "true geometry line" ; or
"[geometric] ray".}
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So: We have:
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→ " [tex]m[/tex]∠[tex]TPQ[/tex] + [tex]m[/tex]∠[tex]QPR[/tex] = 180 " ;
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Given: " [tex]m[/tex]∠[tex]TPQ[/tex] = (11x + 14) " ;
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Plug in this value for: " [tex]m[/tex]∠[tex]TPQ[/tex] " ;
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→ " (11x + 14) + [tex]m[/tex]∠[tex]QPR[/tex] = 180 " ; Solve for: " [tex]m[/tex]∠[tex]QPR[/tex] " :
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→ Subtract: "(11x + 14)" ; from Each Side of the equation:
to isolate: " [tex]m[/tex]∠[tex]QPR[/tex] " ; on one side of the equation;
& to solve for: " [tex]m[/tex]∠[tex]QPR[/tex] " ; in terms of "x" :
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→ " (11x + 14) + [tex]m[/tex]∠[tex]QPR[/tex] − (11x + 14) = 180 − (11x + 14) " ;
Note: On the "left-hand side" of the equation:
The: "(11x + 14)" 's cancel out to "0" ;
{since: (11x + 14) − (11x + 14) = 0 ;
→ {i.e. any value, minus that same value, equals: "zero".}.
→ And we have: " [tex]m[/tex]∠[tex]QPR[/tex] = 180 − (11x + 14) " ;
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Note: " 180 − (11x + 14) " ;
↔ Treat as:
" 180 − 1 (11x + 14) " ;
→ {since multiplying by "1" results in the same value.}.
Consider the following portion:
" − 1 (11x + 14) " ;
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Note the "distributive property" of multiplication:
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→ [tex]a(b +c) = ab + ac[/tex] ;
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Likewise: " − 1 (11x + 14) = (-1*11x) + (-1 *14) ;
= (-11x) + ( -14) '
= - 11x − 14 ;
{Since: "Adding a negative" results in the same value as:
"Subtracting a positive."}.
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Now, bring down the "180" ; and rewrite the expression:
→ " 180 − 11x − 14 " ;
→ Combine the "like terms" :
+ 180 − 14 = + 166 ;
Rewrite the expression as:
→ " 166 − 11x " ;
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Now, we can rewrite the entire equation:
" [tex]m[/tex]∠[tex]QPR[/tex] = 166 − 11x " ;
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Now, consider the triangle: ΔQRP ;
with its 3 (three) sides—as shown in the image attached:
Note:
By definition:
All triangles:
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1) have 3 (three) sides;
2) have 3 (three) angles; and:
3) have angles in which the sum of the measurements of those angles add up to 180°.
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So: For ΔQRP ; which is shown in the image attached:
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Let us consider the measurements of Each of the 3 (three) angles of that triangle:
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1) m∡Q = "(5x + 18)" ; (given);
2) m∡R = " 56 " ; (given) ;
3) m∡P —[within the triangle] = "(166 − 11x)" ; (calculated above}.
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We want to find the value for "x" ;
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So: since all triangles, by definition; have 3 (three) angles with measurements that add up to 180° ;
→ Let us add up the measurements of each of the 3 (three) angles of:
ΔQRP ; and make an equation by setting this sum "equal to: 180 ."
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→ " m∡Q + m∡R + m∡P = 180 " ;
→ (5x + 18) + 56) + (166 - 11x) = 180 ;
→ 5x + 18 + 56 + 166 − 11x = 180 ;
On the "left-hand side" of the equation:
Combine the "like terms" to simplify further:
+5x − 11x = − 6x ;
+ 18 + 56 + 155 = 240 ;
And rewrite the equation:
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- 6x + 240 − 240 = 180 − 240 ;
to get:
- 6x = - 60 ;
Now, divide Each side of the equation by: "( -6)" ;
to isolate: "x" on one side of the equation;
& to solve for "x" ;
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- 6x / 6 = - 60 / -6 ;
to get:
" x = 10 " .
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Now, the question asks for:
" the measurement for angle TPQ ";
→ {that is; " [tex]m[/tex]∠[tex]TPQ[/tex] " } ;
→ which is: " (11x + 14) " ;
Since we know that: " x = 10 " ;
We can plug in our "10" as our value for "x" ; and solve accordingly:
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→ " [tex]m[/tex]∠[tex]TPQ[/tex] = (11x + 14) = (11*10) + 14 = 110 + 14 = 124 .
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The answer is: " [tex]m[/tex]∠[tex]TPQ[/tex] = 124° " .
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Hope this answer—and explanation—is helpful!
Best wishes!
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