Question

Practice set 1( mathamatics) Important for jee mains exam ​

Answer 1

EXPLANATION.

AB is a vertical pole and C is the middle point.

End A is on the level ground.

P at any point on the level ground other then A.

Portion CB subtends an angle β at P.

⇒ AP : AB = 2 : 1.

As we know that,

⇒ AP/AB = 2/1.

⇒ AP = 2AB.

Let we assume that,

Height of AC & AC = h.

⇒ AB = AC + CB.

⇒ AB = h + h = 2h.

⇒ AP = 2(2h) = 4h.

⇒ AP = 4h.

In ΔAPB.

⇒ tanθ = perpendicular/base = p/b.

⇒ tanα = h/4h = 1/4.

⇒ tan(α + β) = 2h/4h.

⇒ tan(α + β) = 1/2.

As we know that,

Formula of :

⇒ tan(α + β) = tanα + tanβ/1 - tanα.tanβ.

⇒ tanα + tanβ/1 - tanα.tanβ = 1/2.

⇒ 1/4 + tanβ/1 - 1/4.tanβ = 1/2.

⇒ 1 + 4tanβ/4/4 - tanβ/4 = 1/2.

⇒ 1 + 4tanβ/4 - tanβ = 1/2.

⇒ 2(1 + 4tanβ) = 4 - tanβ.

⇒ 2 + 8tanβ = 4 - tanβ.

⇒ 8tanβ + tanβ = 4 - 2.

⇒ 9tanβ = 2.

⇒ tanβ = 2/9.

⇒ β = tan⁻¹(2/9).

Option [D] is correct answer.

Answer 2

Answer:

let ∠APC be α

Given :-

∠BPC = β

and

AP:AB = 2:1

$→ \frac{AP}{AB } = \frac{2}{1}$

→ AP = 2AB

C is the midpoint of AB.

let AC = BC = x

then,

AB = 2x and AP = 4x

Now, In ΔCAP,

$\tan \alpha = \frac{AC}{AP}$

$→ \tan \alpha = \frac{x}{4x}$

$→ \tan \alpha = \frac{1}{4}$

again, In ΔBAP,

$\tan( \alpha + \beta ) = \frac{?}{?}$

$→ \tan( \alpha + \beta ) = \frac{2x}{4x}$

$→ \frac{ \tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta } = \frac{1}{2}$

$→2( \tan \alpha + \tan \beta ) = 1 - \tan \alpha \tan \beta$

$→ \tan( \alpha ) + 2 \tan( \beta ) = 1 - \tan( \alpha ) \tan( \beta )$

$→ 2 \tan \beta + \tan \alpha \tan \beta = 1 - 2 \tan \alpha$

$→ \tan \beta (2 + \tan \alpha ) = 1 - 2 \tan \alpha$

putting the value of tanα in the above eqn:-

$→ \tan \beta = \frac{1 - 2 \times \frac{1}{4} }{2 + \frac{1}{4} }$

$→ \tan \beta = \frac{\frac{1}{2} }{ \frac{9}{4} }$

$→ \tan \beta = \frac{2}{9}$

$→ \beta = { \tan }^{ - 1} ( \frac{2}{9} )$