Question

pls answer this question its urgent ​

Answer 1

➼ Answer ➼

(1) $Cos$ $A=\frac{B}{H} =\frac{9}{41}$

Pythagoras Theorem Used:

$H^{2}=P^{2}+B^{2}$

$=>41^{2}=P^{2}+9^{2}$

$=>41^{2}-9^{2}=P^{2}$

$=>1681-81=P^{2}$

$=>1600=P^{2}$

$=>\sqrt{1600} =P$

$=>40=P$

Now,

$\frac{1}{sin^{2}A}-cot^{2}A$

$=>cosec^{2}A -cot^{2}A$

$=>(\frac{H}{P})^{2}-(\frac{B}{P} )^{2}$

$=>(\frac{41}{40})^{2}-(\frac{9}{40} )^{2}$

$=>\frac{1600}{1600}$

$=>1$

(2) Given,

$(2cos2A-1)(tan3A- 1)=0$

Hence,

$2cos$ $2A-1=0$

$=> cos2A=\frac{1}{2}$

$=>cos2A=cos60$

$=>2A=60$

$=>A=30$

Or,

$tan3A-1=0$

$=>tan3A=1$

$=>tan3A=tan45$

$=>3A=45$

$=>A=15$

Thus,

$A=30,$ $15$

☆ Hope it Helps ☆