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Dvinal [7]
3 years ago
5

%7D%7B4%7D%20%7D%20%20%5Ctimes%20%28%28%20%5Cfrac%7B4%7D%7B25%7D%29%20%7B%7D%5E%7B%20%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%5Cdiv%20%28%20%5Cfrac%7B5%7D%7B2%7D%20%29%20%7B%7D%5E%7B%20-%203%7D%20" id="TexFormula1" title="simplify \: ( \frac{81}{16} ) { }^{ \frac{ - 3}{4} } \times (( \frac{4}{25}) {}^{ \frac{3}{2} } \div ( \frac{5}{2} ) {}^{ - 3} " alt="simplify \: ( \frac{81}{16} ) { }^{ \frac{ - 3}{4} } \times (( \frac{4}{25}) {}^{ \frac{3}{2} } \div ( \frac{5}{2} ) {}^{ - 3} " align="absmiddle" class="latex-formula">

Mathematics
1 answer:
BartSMP [9]3 years ago
8 0
Answer as seen in attached.

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Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA
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1. Start with ΔCIJ.

  • ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
  • the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
  • ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

  • ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
  • ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
  • the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
  • ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.

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