Question

1. A tree is broken by the wind . The top struck the ground at an angle of 30° and at a dtance of 30 m from the root. Find the whole heght of the tree.

2. from the top of a hil, the angles of depressio of two consecutive km stones due east are found to be 30° and 45°. Find heght of the hill?

Plz friends solve ​

Answer 1

Step-by-step explanation:

1)Let AB be the tree broken at a point C such that the broken part CB takes the position CO and strikes the ground at O. It is given that OA = 30 m and ∠AOC = 30°.

Answer 2

1.

Correct question :-

• A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 30 m from the root. Find the whole height of the tree.

Solution:-

Let PS be the broken tree at a point R. Broken tree from point R be QR . The top striking the ground at Q.

Then, PQ = 30 m.

And Angle PQR = 30°.

Let, a be RS

and, b be PR.

QR = b

Now, In ∆QPR,

tan 30° = PR/PQ

$\longrightarrow \sf \cfrac{1}{\sqrt{3}} = \cfrac{b}{30}$

• Cross multiple.

$\longrightarrow \sf 1 \times 30 = b \times \sqrt{3}$

$\longrightarrow \sf 30 = b \times \sqrt{3}$

$\longrightarrow \sf \cfrac{30}{\sqrt{3}} = b$

• Rationalise the denominator.

$\longrightarrow \sf \cfrac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = b$

$\longrightarrow \sf \cfrac{30 \times \sqrt{3}}{3} = b$

$\longrightarrow \sf 10 \sqrt{3} = b$

In ∆QPR,

cos 30° = PQ/QR

$\longrightarrow \sf \cfrac{\sqrt{3}}{2} = \cfrac{30}{a}$

• Cross multiple.

$\longrightarrow \sf a \times \sqrt{3} = 30 \times 2$

$\longrightarrow \sf a \times \sqrt{3} = 60$

$\longrightarrow \sf a = \cfrac{60}{\sqrt{3}}$

• Rationalise the denominator.

$\longrightarrow \sf a = \cfrac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$\longrightarrow \sf a = \cfrac{60 \times \sqrt{3}}{3}$

$\longrightarrow \sf a = 20 \sqrt{3}$

Height of tree = a + b

= 20√3 + 10√3

= 30√3

= 30 × 1.73

= 51.96

Therefore,

Height of tree is 51.96 m.

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2.

Correct question :-

• From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. Find height of the hill?

Solution:-

Let, PQ be the hill. PQ is of height of a km.

R and S be two stones.

Angle of depression of S and R be 45° and 30° respectively.

Distance of QS be b km.

In ∆SQP

tan 45° = PQ/QS

$\longrightarrow \sf 1 = \cfrac{a}{b}$

$\longrightarrow \sf b = a$ -----(1)

In ∆RQP

tan 30° = PQ/QR

$\longrightarrow \sf \cfrac{1}{\sqrt{3}} = \cfrac{a}{b+1}$

$\longrightarrow \sf b+1 = a \times \sqrt{3}$

• By equation (i) b = a. So, Puting value of b be a .

$\longrightarrow \sf a + 1 = a \times \sqrt{3}$

$\longrightarrow \sf 1 = a(\sqrt{3} - 1)$

$\longrightarrow \sf \cfrac{1}{\sqrt{3} - 1} = a$

• Rationalise the denominator.

$\longrightarrow \sf \cfrac{1 \times ( \sqrt{3} + 1)}{(\sqrt{3} -1 )( \sqrt{3} + 1)} = a$

$\longrightarrow \sf \cfrac{\sqrt{3}+ 1}{(\sqrt{3})^{2} - (1)^{2}} = a$

$\longrightarrow \sf \cfrac{\sqrt{3} + 1}{3 - 1} = a$

$\longrightarrow \sf \cfrac{\sqrt{3} + 1}{2} = a$

$\longrightarrow \sf \cfrac{1.73 + 1}{2} = a$

$\longrightarrow \sf \cfrac{2.73}{2} = a$

$\longrightarrow \sf 1.36 = a$

We have taken a be height of hill.

Therefore,

The height of hill is 1.36 km .

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Diagrams of both questions are in attachment.