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Consider a triangle ABC like the one below. Suppose that =A110°, =b25, and =c4. (The figure is not drawn to scale.) Solve the triangle.
Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

If there is more than one solution, use the button labeled "or".

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nuhulawal20

Side a, angle B and angle C of triangle ABC are 26.6 units, 61.9 degrees and 8.1 degrees respectively.

What is the value of the missing angles and side?

Given that;

  • Angle A = 110°
  • Side b = 25
  • side c = 4
  • side a = ?
  • Angle B = ?
  • Angle C = ?

First we determine the length of side a

[tex]a = \sqrt{b^2-c^2-2bc*cos(A)}[/tex]

We substitute our values into the above equation[tex]a = \sqrt{25^2-4^2-(2*25*4*cos(110)}\\\\a = 26.6[/tex]

Side a has a length of 26.6 units.

[tex]B = cos^{-1}(\frac{a^2+c^2-b^2}{2ac}) \\\\B = cos^{-1}(\frac{26.63464^2+4^2-25^2}{2*26.63464*4}) \\\\B = 61.9[/tex]

Angle B is 61.9 degrees.

[tex]C = cos^{-1}(\frac{a^2+b^2-c^2}{2ab} )\\\\C = cos^{-1}(\frac{26.63464^2+25^2-4^2}{2*26.63464*25} )\\\\C =8.1[/tex]

Angle C is 8.1 degrees.

Therefore, side a, angle b and angle c of the triangle ABC are 26.6 units, 61.9 degrees and 8.1 degrees respectively.

Learn more about triangles here: https://brainly.com/question/14882091

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