Question

QUESTION :

The elevation of the top of a temple from a point A situated in south is 45° and B is a point to the west of the point A. Also if the elevation of the top of the temple from B is 15° and AB = 2a, then find the height of the temple.

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Answer 1

ANSWER :–

Height of the temple = h = a[2/√(6 + 4√3)]

EXPLANATION :–

GIVEN :–

• The elevation of the top of a temple from a point A situated in south is 45° .

• B is a point to the west of the point A.

• elevation of the top of the temple from B is 15°.

• Distance (AB) = 2a

TO FIND :–

Height of the temple.

SOLUTION :–

• Let's assume the height of the temple = h

⮕ in △ APO –

=> tan(45⁰) = h/(AP)

=> AP = h

⮕ in △ BPO –

=> tan(15⁰) = h/(BP)

=> 2 - √3 = h/(BP)

=> BP = h/(2 - √3)

=> BP = [h/(2 - √3)][(2 + √3)/(2 + √3)

=> BP = h(2 + √3)

⮕ Now , Applying pythagoras theorm in △ PAB –

=> (BP)² = (AP)² + (AB)²

=> [h(2 + √3)]² = h² + (2a)²

=> h²(2 + √3)² = h² + 4a²

=> h²(4 + 3 + 4√3) = h² + 4a²

=> h²(7 + 4√3) = h² + 4a²

=> h²(7 + 4√3 - 1) = 4a²

=> h²(6 + 4√3) = 4a²

=> h² = (4a²)/(6 + 4√3)

=> h = a[2/√(6 + 4√3)]

Answer 2

Given:

• The elevation of the top of a temple from a point A situated in south is 45°
• B is a point to the west of the point A
• The elevation of the top of the temple from B is 15°
• Length of AB= 2a

To Find:

Height of the temple

Solution:

Let CD be the temple of height 'x'.

Diagram(See attachment 1)

According to diagram, A is the point situated at south of the temple and B is the point which is west to the point A  

In ΔACD

Diagram(See attachment 2)

$\longrightarrow\mathrm{tan45^{\circ}=\dfrac{CD}{AD}}$

$\longrightarrow\mathrm{1=\dfrac{x}{AD}}$

$\longrightarrow\mathrm{AD=x}$

Now, in ΔABD

Diagram(See attachment 3)

On using Pythagoras Theorem, we get

$\longrightarrow\mathrm{(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}}$

$\longrightarrow\mathrm{(BD)^{2}=(AB)^{2}+(AD)^{2}}$

$\longrightarrow\mathrm{(BD)^{2}=(2a)^{2}+(x)^{2}}$

$\longrightarrow\mathrm{(BD)^{2}=4a^{2}+x^{2}}$

$\longrightarrow\mathrm{BD=\sqrt{4a^{2}+x^{2}}}$

Now, in ΔBCD

Diagram(See attachment 4)

$\longrightarrow\mathrm{tan15^{\circ}=\dfrac{CD}{BD}}$

$\longrightarrow\mathrm{2-\sqrt{3}=\dfrac{x}{\sqrt{4a^{2}+x^{2}}}}$

On squaring both the sides, we get

$\longrightarrow\mathrm{(2-\sqrt{3})^{2}=\dfrac{x^{2}}{4a^{2}+x^{2}}}$

$\longrightarrow\mathrm{(2)^{2}+(\sqrt{3})^{2}-2(2)(\sqrt{3})=\dfrac{x^{2}}{4a^{2}+x^{2}}}$

$\longrightarrow\mathrm{4+3-4\sqrt{3}=\dfrac{x^{2}}{4a^{2}+x^{2}}}$

$\longrightarrow\mathrm{7-4\sqrt{3}=\dfrac{x^{2}}{4a^{2}+x^{2}}}$

$\longrightarrow\mathrm{4a^{2}+x^{2}=\dfrac{x^{2}}{7-4\sqrt{3}}}$

On rationalising R.H.S., we get

$\longrightarrow\mathrm{4a^{2}+x^{2}=\dfrac{x^{2}}{7-4\sqrt{3}}\times \dfrac{7+4\sqrt{3}}{7+4\sqrt{3}}}$

$\longrightarrow\mathrm{4a^{2}+x^{2}=\dfrac{x^{2}(7+4\sqrt{3})}{(7)^{2}-(4\sqrt{3})^{2}}}$

$\longrightarrow\mathrm{4a^{2}+x^{2}=\dfrac{x^{2}(7+4\sqrt{3})}{49-48}}$

$\longrightarrow\mathrm{4a^{2}+x^{2}=\dfrac{7x^{2}+4\sqrt{3}x^{2}}{1}}$

$\longrightarrow\mathrm{4a^{2}+x^{2}=7x^{2}+4\sqrt{3}x^{2}}$

$\longrightarrow\mathrm{4a^{2}=7x^{2}-x^{2}+4\sqrt{3}x^{2}}$

$\longrightarrow\mathrm{4a^{2}=6x^{2}+4\sqrt{3}x^{2}}$

$\longrightarrow\mathrm{4a^{2}=x^{2}(6+4\sqrt{3})}$

$\longrightarrow\mathrm{x^{2}(6+4\sqrt{3})=4a^{2}}$

$\longrightarrow\mathrm{x^{2}=\dfrac{4a^{2}}{6+4\sqrt{3}}}$

$\longrightarrow\mathrm{x=\sqrt{\dfrac{4a^{2}}{6+4\sqrt{3}}}}$

$\longrightarrow\mathrm{x=\dfrac{2a}{\sqrt{6+4\sqrt{3}}}}$

Hence, height of the temple is $\dfrac{2a}{\sqrt{6+4\sqrt{3}}}$