Question

exterior angle sum of triangle with 4.9 cm,5.9 cm 6.8 cm . urgently needed ,
draw the triangle with the cm .
then, measure the angles and tell. ​

Answer 1

Step-by-step explanation:

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\begin{gathered}so \: here \: \\ secA + tanA =x \\ ie \: tan \: A = x - sec \: A\end{gathered}

sohere

secA+tanA=x

ietanA=x−secA

\begin{gathered}so \: now \: using \\ 1 + tan ^{2} A = sec {}^{2} A \\ 1 + (x - sec \: A) {}^{2} = sec {}^{2} A \\ 1 + x {}^{2} + sec {}^{2} A - 2sec \: A.x = sec {}^{2} A \\ 1 + x {}^{2} - 2sec \: A.x = 0 \\ ie \: 1 + x {}^{2} = 2secA.x \\ sec \: A = 1 + x {}^{2} /2x \\ \\ \end{gathered}

sonowusing

1+tan

2

A=sec

2

A

1+(x−secA)

2

=sec

2

A

1+x

2

+sec

2

A−2secA.x=sec

2

A

1+x

2

−2secA.x=0

ie1+x

2

=2secA.x

secA=1+x

2

/2x

\begin{gathered}thus \: value \: of \: sec \: A \: (in \: terms \: of \: x) \: is \: 1 + x {}^{2} /2x \\ \end{gathered}

Answer 2

Answer:

4.9 + 5.9 + 6.8

= 17.6

Exterior angle sum of triangle = 17.6

Hope it will help you

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