explain step by step how to solve. please point out any similarities in the steps used for each question

Mathematics

1 answer:

Mamont248 [21]3 years ago

6 0

#1
m∠A=180 - 115 - 24 = 41°

By the law of sines:

b/sinB = a/sinA ⇒
b = (a*sinB)/sinA = (21*sin24°)/sin41° = (21*0.4067)/0.656 ≈ 13

c/sinC = a/sinA ⇒
c = (a*sinC)/sinA = (21*sin115°)/sin41° = (21*0.9063)/0.656 ≈ 29

#2
m∠C=180 - 119 - 27 = 34°

By the law of sines:

b/sinB = a/sinA ⇒
b = (a*sinB)/sinA = (13*sin119°)/sin27° = (13*0.8746)/0.454 ≈ 25

c/sinC = a/sinA ⇒
c = (a*sinC)/sinA = (13*sin34°)/sin27° = (13*0.5592)/0.454 ≈ 16

#3
m∠C=180 - 57 - 37 = 86°

By the law of sines:

c/sinC = a/sinA ⇒
c = (a*sinC)/sinA = (11*sin86°)/sin57° = (11*0.9976)/0.8387 ≈ 13.1

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What is equivalent decimal for 0.85

Alex_Xolod [135]

Answer:

17/20

Step-by-step explanation:

0.85 = 85/100 = 17/20

The number 0.85 can be written using the fraction 85/100 which is equal to 17/20 when reduced to lowest terms.

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inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$