contestada

As part of calculations to solve an oblique plane triangle (ABC), the following data was available: b=50.071 horizontal distance, C=90.286° (decimal degrees), B=62.253° (decimal degrees). Calculate the distance of c to 3 decimal places (no alpha).

Respuesta :

Answer:

The distance of c is 56.57

Explanation:

Given that,

Horizontal distance b = 50.071

Angle C = 90.286°

Angle B = 62.253°

We need to calculate the distance of c

Using sine rule

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

Put the value into the formula

[tex]\dfrac{50.071}{\sin62.253 }=\dfrac{c}{\sin90.286}[/tex]

[tex]c= \dfrac{50.071\times\sin90.286}{\sin62.253}[/tex]

[tex]c=56.575[/tex]

Hence, The distance of c is 56.575.