Question

If ABCD is a trapezium in which AB || DC, AB = 15cm, BC = 6cm, CD = 7cm and AD = 4cm.
2 COSA + 3 cosB =
[ ]​

Answer 1

Given:

• AB || DC
• AB = 15cm
• BC = 6cm
• CD = 7cm
• AD = 4cm

Find:

• Value of 2 cos A + 3 cos B = ?

Solution:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \thicklines \put(2,2){\line(1,2){1.5}} \put(3.5,2){\line(0,1){3}} \put(2,2){\line(1,0){1.5}} \put(3.7,2){\line(0,1){0.2}} \put(3.5,2.2){\line(1,0){0.2}} \put(3.5,5){\line(1,0){3}} \put(6.5,2){\line(0,1){3}}\put(9.5,2){\line( - 1,1){3}}\put(3.5,2){\line(1,0){6}} \put(6.8,2){\line(0,1){0.2}} \put(6.5,2.2){\line(1,0){0.3}} \put(9.2,1.6){\sf B} \put(2,1.6){\sf A} \put(3.4,1.6){\sf P} \put(6.4,1.6){\sf Q} \put(3.4,5.2){\sf D} \put(6.4,5.2){\sf C} \put(5,1.4){\vector(1,0){3}} \put(6,1.4){\vector( - 1,0){3}} \put(5,1){\sf 15cm} \put(1.8,3.1){\sf 4cm} \put(8.2,3.4){\sf 6cm} \put(4.5,5.2){\sf 7cm} \end{picture}$

In $\triangle$ ADP,

$\sf \green\star \cos A = \dfrac{B}{H} \\ \\ \sf \pink\star \cos A = \dfrac{AP}{AD}$

where,

• AD = 4cm

So,

$\sf \orange\star \cos A = \dfrac{AP}{4}......1$

Now, In $\triangle$ QBC

$\sf \pink\star \cos B = \dfrac{B}{H} \\ \\ \sf \red\star \cos B = \dfrac{QB}{CB}$

where,

• CB = 6cm

So,

$\sf \blue\star \cos B = \dfrac{QB}{6}.......2$

Use eq 1 and 2 in Question

$\sf \dashrightarrow 2 \cos A + 3 \cos B \\ \\ \\ \sf \dashrightarrow 2 \big\lgroup {\dfrac{AP}{4}} \big\rgroup + 3 \big\lgroup {\dfrac{QB}{6}} \big\rgroup \\ \\ \\ \sf \dashrightarrow \dfrac{AP}{2} + \dfrac{QB}{2} \\ \\ \\ \sf \dashrightarrow \dfrac{ AP + QB }{2} \\ \\ \big\lgroup{ \sf we, \: can \: write \: AP + QB = AB - PQ } \big\rgroup \\ \\ \\ \sf \dashrightarrow \dfrac{ AB - PQ }{2} \\ \\ \big\lgroup{ \sf PQ = CD } \big\rgroup \qquad \qquad \{ \sf By \: Construction \} \\ \\ \sf \star So, \\ \\ \\ \sf \dashrightarrow \dfrac{ AB - PQ }{2} =\dfrac{ AB - CD }{2}$

where,

• CD = 7cm
• AB = 15cm

So,

$\sf \dashrightarrow \dfrac{ AB - CD }{2} \\ \\ \\ \sf \dashrightarrow \dfrac{ 15 - 7}{2} = \dfrac{8}{2} \\ \\ \\ \sf \dashrightarrow 4$

Hence, 2 cos A + 3 cos B = 4