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7. if (mangle lmp) is 11 degrees more than (mangle nmp) and (mangle nml…

Question

  1. if (mangle lmp) is 11 degrees more than (mangle nmp) and (mangle nml = 137^{circ}), find each measure.
  2. if (mangle abc) is one degree less than three times (mangle abd) and (mangle dbc = 47^{circ}), find each measure.
  3. if (overline{qs}) bisects (angle pqt), (mangle sqt=(8x - 25)^{circ}), (mangle pqt=(9x + 34)^{circ}), and (mangle sqr = 112^{circ}), find each measure.
  4. if (angle cde) is a straight angle, (overline{de}) bisects (angle gdh), (mangle gde=(8x - 1)^{circ}), (mangle edh=(6x + 15)^{circ}), and (mangle cdf = 43^{circ}), find each measure.

Explanation:

Step1: Solve problem 8

Let $m\angle ABD = x$. Then $m\angle ABC=3x - 1$. Since $m\angle ABC=m\angle ABD + m\angle DBC$ and $m\angle DBC = 47$, we have the equation $3x-1=x + 47$.
Subtract $x$ from both sides: $3x-1-x=x + 47-x$, which simplifies to $2x-1=47$.
Add 1 to both sides: $2x-1 + 1=47+1$, so $2x=48$.
Divide both sides by 2: $x = 24$. So $m\angle ABD=24^{\circ}$ and $m\angle ABC=3\times24-1=71^{\circ}$.

Step2: Solve problem 9

Since $\overline{QS}$ bisects $\angle PQT$, then $m\angle PQT = 2m\angle SQT$. So $9x + 34=2(8x - 25)$.
Expand the right - hand side: $9x + 34=16x-50$.
Subtract $9x$ from both sides: $9x + 34-9x=16x-50-9x$, which gives $34 = 7x-50$.
Add 50 to both sides: $34 + 50=7x-50+50$, so $84 = 7x$.
Divide both sides by 7: $x = 12$.
$m\angle PQS=m\angle SQT=8\times12-25=96 - 25 = 71^{\circ}$, $m\angle PQT=9\times12 + 34=108+34 = 142^{\circ}$, $m\angle TQR=112 - 71 = 41^{\circ}$.

Step3: Solve problem 10

Since $\overline{DE}$ bisects $\angle GDH$, then $m\angle GDE=m\angle EDH$. So $8x-1=6x + 15$.
Subtract $6x$ from both sides: $8x-1-6x=6x + 15-6x$, which gives $2x-1=15$.
Add 1 to both sides: $2x-1 + 1=15+1$, so $2x=16$.
Divide both sides by 2: $x = 8$.
$m\angle GDH=2(8\times8 - 1)=2\times63 = 126^{\circ}$.
Since $\angle CDE$ is a straight angle ($180^{\circ}$) and $m\angle CDF = 43^{\circ}$, then $m\angle FDE=180 - 43=137^{\circ}$, $m\angle FDH=m\angle FDE - m\angle EDH$. $m\angle EDH=6\times8+15=48 + 15=63^{\circ}$, so $m\angle FDH=137-63 = 74^{\circ}$.

Answer:

  1. $m\angle ABD = 24^{\circ}$, $m\angle ABC = 71^{\circ}$
  2. $x = 12$, $m\angle PQS = 71^{\circ}$, $m\angle PQT = 142^{\circ}$, $m\angle TQR = 41^{\circ}$
  3. $x = 8$, $m\angle GDH = 126^{\circ}$, $m\angle FDH = 74^{\circ}$, $m\angle FDE = 137^{\circ}$