Question

if cos (a+b)=4/5 and sin(a-b)=5/13.when 0<a,b<=45°,then tan 2a is​

Answer 1

${\mathcal{\red{ANSWER}}}$

Given:

$\tt{\cos\;(a+b)=\dfrac{4}{5}\;and\;\sin\;(a-b)=\dfrac{5}{13}}$

$\tt{\implies \cos\;(a+b)=\dfrac{4}{5}=\dfrac{B}{H} \;and\;\sin\;(a-b)=\dfrac{5}{13}}$

$\tt{\implies \sin\;(a+b)=\dfrac{3}{5} =\dfrac{P}{H} \;\;[use\;Pythagoras\;theorem\;to\;find\;perpendicular]}$

$\tt{\implies (a+b)=\sin^{-1}\dfrac{3}{5} \;\;\;.......(1)}$

$\tt{and\;(a-b)=sin^{-1}\dfrac{5}{13}\;\;\;......(2)}$

On adding equation (1) and (2) we get,

$\tt{\implies 2a=\sin^{-1}\dfrac{3}{5} +\sin^{-1}\dfrac{5}{13}}$

$\tt{\implies 2a = \sin^{-1}\Bigg[\dfrac{3}{5}\sqrt{1-\bigg(\dfrac{5}{13}\bigg)^{2}}+\dfrac{5}{13}\sqrt{1-\bigg(\dfrac{3}{5}\bigg)^{2}}\Bigg]}$

$\implies 2a=\sin^{-1}\Bigg[\dfrac{3}{5}\sqrt{\dfrac{144}{169}}+\dfrac{5}{13}\sqrt{\dfrac{16}{25}}\Bigg]$

$\tt{\implies 2a=\sin^{-1}\Bigg[\dfrac{3}{5} \times \dfrac{12}{13} +\dfrac{5}{13} \times \dfrac{4}{5}\Bigg]=\sin^{-1}\Bigg[\dfrac{56}{65}\Bigg]}$

$\tt{\implies \sin 2a=\dfrac{56}{65} = \dfrac{P}{H}}$

$\tt{Now,\;B^{2}=H^{2}-P^{2}=65^{2}-56^{2}=4225-3136=1089}$

$\tt{\implies B=\sqrt{1089}=33}$

$\tt{Now,\;\tan 2a = \dfrac{P}{B}=\dfrac{56}{33}}$

_____________________________________________[ANSWER]

Answer 2

Step-by-step explanation:

maximum height can be attained only when the projectile is launched at 90°

threrfore ,

h= u^2/2g

25×2×9.8= u^2

490=u^2

7√10=u

then for range ,

the maximum range occurs at 45°

R = u^2/g

R = 490/9.8

R = 50 m