calculate the value of a,a and c in d ABC given that b=17.23cm,c=10.86cm and b=101.3

You have typed something in incorrectly.

A triangle can't have two different values for b.
Are there two triangles? Or is one suppose be the angle versus the side length?
Please advise.

One is capital letter while the other is small letter b

To calculate the values of angle A, angle B, and side a in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c:

c^2 = a^2 + b^2 - 2ab*cos(C)

In this case, we are given that b = 17.23 cm, c = 10.86 cm, and angle C (opposite side c) is 101.3 degrees. We need to solve for a, angle A, and angle B.

Step 1: Apply the Law of Cosines
c^2 = a^2 + b^2 - 2ab*cos(C)
(10.86)^2 = a^2 + (17.23)^2 - 2*a*(17.23)*cos(101.3)

Step 2: Solve for a
Rearrange the equation to solve for a:
a^2 = (10.86)^2 + (17.23)^2 - 2*a*(17.23)*cos(101.3)
a^2 - 2*a*(17.23)*cos(101.3) = (10.86)^2 + (17.23)^2
a^2 - 2*a*17.23*(-0.2926) = (10.86)^2 + (17.23)^2
a^2 + 10.0486*a + 517.72 = 118.3396 + 297.3129
a^2 + 10.0486*a - 702.7529 = 0

This is now a quadratic equation. Solve it using any quadratic formula or calculator to find the values of a.

Step 3: Calculate angles A and B
Using the Law of Sines:
sin(A)/a = sin(C)/c

We already know c and we will now have the value of a from solving the quadratic equation. So, we can substitute these values into the equation to solve for angle A.

sin(A)/a = sin(C)/c
sin(A)/a = sin(101.3)/10.86

Now, rearrange the equation to solve for sin(A):
sin(A) = (sin(101.3)/10.86) * a

Use the inverse sine (sin^(-1)) function to find angle A:
A = sin^(-1)((sin(101.3)/10.86) * a)

Similarly, we can calculate angle B using the equation:
B = 180 - A - C

After substituting the calculated values, we can find angle B.