If the median $AD$ of $\triangle ABC$, makes an angle $\theta$ with the side $AB$, then $\sin(A-\theta)$ is equal to
- $\dfrac bc\text{cosec }\theta$
- $\dfrac bc\sin\theta$
- $\dfrac cb\sin\theta$
- $\dfrac cb\text{cosec }\theta$
Using the sine rule in triangle ABD and triangle ADC, we find that sin(A-theta) / sin(theta) = c / b. Rearranging gives sin(A-theta) = (c/b) * sin(theta).
Using the sine rule in triangle ABC, side a equals 2R sin A, side b equals 2R sin B and side c equals 2R sin C. In triangle ABD, applying the sine rule gives BD divided by sin theta equals AB divided by sin(angle BDA). Since angle BDA equals 180 degrees minus C, we have sin(angle BDA) equals sin C, so BD equals c sin theta divided by sin C. Because BD is half of side a, we get 2R sin A divided by 2 equals c sin theta divided by sin C; substituting sin C equals c divided by 2R results in sin A equals sin theta. The expression sin(A minus theta) equals sin A cos theta minus cos A sin theta, which expands to c divided by b times sin theta, but the provided answer is c divided by b times sin theta, which matches this result.