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Setler79 [48]
3 years ago
13

10. 10. If AB bisects CAF, and m<EAF = 72°,then the m<BAF =​

Mathematics
1 answer:
never [62]3 years ago
4 0

The measure of ∠BAF is 54°.

Solution:

DF and CE are intersecting lines.

m∠EAF = 72° and AB bisects ∠CAF.

∠EAF and ∠DAC are vertically opposite angles.

Vertical angle theorem:

<em>If two lines are intersecting, then vertically opposite angles are congruent.</em>

∠DAC ≅ ∠EAF

m∠DAC = 72°

<em>Sum of the adjacent angles in a straight line = 180°</em>

m∠DAE + m∠EAF = 180°

m∠DAE + 72° = 180°

Subtract 72° from both sides.

m∠DAE = 108°

∠CAF and ∠DAE are vertically opposite angles.

⇒ m∠CAF = m∠DAE

⇒ m∠CAF = 108°

AB bisects ∠CAF means ∠CAB = ∠BAF

m∠CAB + m∠BAF = 108°

m∠BAF + m∠BAF = 108°

2 m∠BAF = 108°

Divide by 2 on both sides, we get

m∠BAF = 54°

Hence the measure of ∠BAF is 54°.

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VARVARA [1.3K]

Answer:

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Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}

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\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

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\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}

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