Answer:
<u>Given:</u>
- DC ║ AB
- CM = MB as M is midpoint of BC
i) <u>Since DN and BC are transversals, we have:</u>
- ∠DCM ≅ ∠NBM and
- ∠CDM ≅ ∠BNM as alternate interior angles
<u>As two angles and one side is congruent, the triangles are also congruent:</u>
- ΔDCM ≅ ΔNBM (according to AAC postulate)
So their areas are same.
ii)
<u>The quadrilateral has area of:</u>
- A(ADCB) = A(ADMB) + A(DCM)
<u>And the triangle has area of:</u>
- A(ADN) = A(ADMB) + A(NBM)
Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.
Cos( A + B ) = cosAcosB - sinAsinB ;
cos( A + B ) / ( cosAsinB ) = ( cosAcosB - sinAsinB ) / ( cosAsinB ) = ( cosAcosB ) / ( cosAsinB ) - ( sinAsinB ) / ( cosAsinB ) = cosB / sinB - sinA / cosA = cotB - tanA ;
Answer:
A.
Step-by-step explanation:
