Step 1 Given, a right angle triangle ∠E=90∘ ED=6 EF=3 Step 2 Let ∠F=θ then tan⁡θ=pb=DEEF ⇒tan⁡θ=63 ⇒tan⁡θ=2 ⇒θ=tan−12 ⇒θ=63.434 =∼63.4 ∴ on ∠F=63.4

Step 1 b=6 c=3 α=90∘ Step 2 a2=b2+c2−2bccos⁡α a=b2+c2−2bccos⁡α a=62+32−2×6×3×cos⁡90∘ a=6.71 Step 3 a=6.71 b=6 c=30 a2=b2+c2−2bccos⁡α α=arccos⁡(b2+c2−a22bc)=arccos⁡(62+32−6.7122×6×3)=90∘ b2=a2+c2−2accos⁡β β=arccos⁡(a2+c2−b22ac)=arccos⁡(6.712+32−622×6.71×3)=63∘ Answer: m∠F=63∘

Step 1 Given, a right angle triangle ∠E=90∘ ED=6 EF=3 Step 2 Let ∠F=θ then tan⁡θ=pb=DEEF ⇒tan⁡θ=63 ⇒tan⁡θ=2 ⇒θ=tan−12 ⇒θ=63.434 =∼63.4 ∴ on ∠F=63.4

Step 1 b=6 c=3 α=90∘ Step 2 a2=b2+c2−2bccos⁡α a=b2+c2−2bccos⁡α a=62+32−2×6×3×cos⁡90∘ a=6.71 Step 3 a=6.71 b=6 c=30 a2=b2+c2−2bccos⁡α α=arccos⁡(b2+c2−a22bc)=arccos⁡(62+32−6.7122×6×3)=90∘ b2=a2+c2−2accos⁡β β=arccos⁡(a2+c2−b22ac)=arccos⁡(6.712+32−622×6.71×3)=63∘ Answer: m∠F=63∘

"Find m∠F" was answered by students like you

Josh Sizemore

Answered question

2021-12-14

Find

m ∠ F

Answer & Explanation

Vivian Soares

Beginner2021-12-15Added 36 answers

Step 1
Given,
a right angle triangle

∠ E = 90^{\circ}

E D = 6

E F = 3

Step 2
Let

∠ F = θ

then

\tan θ = \frac{p}{b} = \frac{D E}{E F}

\Rightarrow \tan θ = \frac{6}{3}

\Rightarrow \tan θ = 2

\Rightarrow θ = \tan^{- 1} 2

\Rightarrow θ = 63.434

\tilde{=} 63.4

∴ o n ∠ F = 63.4

Ethan Sanders

Beginner2021-12-16Added 35 answers

Step 1
$b = 6$
$c = 3$
$α = 90^{\circ}$
Step 2
$a^{2} = b^{2} + c^{2} - 2 b c \cos α$
$a = \sqrt{b^{2} + c^{2} - 2 b c \cos α}$
$a = \sqrt{6^{2} + 3^{2} - 2 \times 6 \times 3 \times {\cos 90}^{\circ}}$
$a = 6.71$
Step 3
$a = 6.71$
$b = 6$
$c = 30$
$a^{2} = b^{2} + c^{2} - 2 b c \cos α$
$α = \arccos (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) = \arccos (\frac{6^{2} + 3^{2} - {6.71}^{2}}{2 \times 6 \times 3}) = 90^{\circ}$
$b^{2} = a^{2} + c^{2} - 2 a c \cos β$
$β = \arccos (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) = \arccos (\frac{{6.71}^{2} + 3^{2} - 6^{2}}{2 \times 6.71 \times 3}) = 63^{\circ}$
Answer: $m ∠ F = 63^{\circ}$

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