Question

Xcos phi1 + ysin phi1 = a and xcos phi2 + ysin phi2 = a find point of intersection of the straight line

Answer 1

I have changed few variables(angles).

Answer:

(a(sinA - sinB)/sin(A - B), a(cosB - cosA)/sin(A - B)) , A = phi1 & B = phi2

Step-by-step explanation:

xcosA + ysinA = a ... (1)

xcosB + ysinB = a ... (2)

Multiply (1) & (2) with sinB & sinA:

=> xcosAsinB + ysinAsinB = asinB

&

=> xcosBsinA + ysinBsinA = asinA

Subtract one from another, here new(1) - new(2):

xcosAsinB + ysinAsinB = asinB

xcosBsinA + ysinBsinA = asinA

- - -

x(cosAsinB-cosBsinA) = a(sinB-sinA)

=> x(cosAsinB-cosBsinA) = a(sinB-sinA)

=> xsin(B - A) = a(sinB - sinA)

=> x = a(sinB - sinA)/sin(B - A)

Or, = a(sinA - sinB)/sin(A - B)

Now, multiply (1) & (2) with cosB & cosA:

xcosAcosB + ysinAcosB =acosB

&

xcosAcosB + ysinBcosA = acosA

Subtract one from another, here new(1) - new(2):

xcosAcosB + ysinAcosB =acosB

xcosAcosB + ysinBcosA = acosA

- - -

y(sinAcosB - sinBcosA) = a(cosB - cosA)

=> y(sin(A - B)) = a(cosB - cosA)

=> y = a(cosB - cosA)/sin(A - B)

Thus, points are:

(x, y) = (a(sinA - sinB)/sin(A - B), a(cosB - cosA)/sin(A - B))