# Solve the triangle. B = 36°, a = 38, c = 17

**Solution:**

The information provided in the problem statement is summarised in the figure below:

By solving the triangle it implies that the values of ∠A , ∠C and side b have to be determined.

Therefore to obtain these values a perpendicular is dropped from vertex A on to the side BC as shown in the diagram below:

We know that.

Cos36° = BD/AB And AB = c = 17

Cos36° = 0.8089

Therefore.

BD = ABcos36°

= 17cos36°

= 17( 0.8089)

= 13.75

Therefore we can now determine DC from the relationship

DC = BC - BD

= 38 - 13.75

= 24.25

We also know sin36° = AD/AB

AB = 17 and sin 36°= 0.588 AD = ABsin36°

= (17)(0.588)

= 9.996

Now

Tan C° = AD/DC

= 9.996/24.25

=0.4122

C° = 22.39

The Cosine of angle C is given as cosC° = DC/AC

C° = 22.39 and DC = 24.25

Therefore

AC = DC/cos22.39°

cos 22.39° = 0.9245

Therefore

AC = 24.25/0.9245

= 26.23

∠A = 180° - 36° - 22.39°

= 121.61°

Therefore all the parameters of the triangle are obtained as follows:

a = 38; c = 17; b = 26.23; ∠A = 121.61°; ∠B = 36° and ∠C = 22.39°

## Solve the triangle. B = 36°, a = 38, c = 17

**Summary:**

Solving the given triangle with ∠B = 36°, a = 38, c = 17 we have b = 26.23; ∠A = 121.61°; and ∠C = 22.39°.

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