Question

A clinometer is a tool that is used to measure the angle of elevation, or angle from the ground, in a right - angled triangle. We can use a clinometer to measure the height of tall things that you can’t possibly reach to the top of, flag poles, buildings, trees. Ravish got a clinometer from school lab and started the measuring elevation angle in surrounding. He saw a building on which society logo is painted on wall of building. From a point P on the ground level, the angle of elevation of the roof of the building is 45 degree. The angle of elevation of the centre of logo is 30 degree from same point. The point P is at a distance of 24 m from the base of the building.
(i) What is the height of the building logo from ground ? (a) 8√2 m (b) 4√3 m (c) 8 √3 m (d) 4√2 m
(ii) What is the height of the building from ground ? (a) 24 (3-√3) m (b) 8 (3-√3) m (c) 24 m (d) 32 m
(iii) What is the aerial distance of the point P from the top of the building ? (a) 24√3 m (b) 24√2 m (c) 32√3 m (d) 32√2 m
(iv) In above case the angle of elevation ϕ of the top of building is given by (a) tan ϕ = 1.6 (b) tan ϕ = 1.5 (c) tanϕ = 0.75 (d) tanϕ = 0.8​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Concept:

tan A = (Opposite side of angle A)/(Adjacent side of angle A)

Given:

A clinometer is a tool that is used to measure the angle of elevation, or angle from the ground, in a right - angled triangle. We can use a clinometer to measure the height of tall things that you can’t possibly reach to the top of, flag poles, buildings, trees. Ravish got a clinometer from school lab and started the measuring elevation angle in surrounding. He saw a building on which society logo is painted on wall of building. From a point P on the ground level, the angle of elevation of the roof of the building is 45 degree. The angle of elevation of the centre of logo is 30 degree from same point. The point P is at a distance of 24 m from the base of the building.

Find:

The height of the building logo from ground, the height of the building from ground, the aerial distance of the point P from the top of the building, In above case if the angle of elevation ϕ then relation if the P move towards base of building 9m.

Solution:

The suitable figure for this problem,

Q is the base of the building, R is roof of the building and L is the center of the logo and P is the point.

PQ = 24 m

So from the right angled triangle PQL,

QL = height of the logo from the ground

Now,

$\tan30^\circ=\frac{QL}{PQ}\\\Rightarrow\frac{1}{\sqrt3}=\frac{QL}{24}\\\Rightarrow QL=\frac{24}{\sqrt3}=8\sqrt3$

From the right angled triangle PQR,

QR = height of the building from the ground.

Now,

$\tan45^\circ=\frac{QR}{PQ}\\\Rightarrow1=\frac{QR}{24}\\\Rightarrow QR=24$

Hypotenuse PR is the aerial dustance of the point P from the top of the building,

By Pythagorean Theorem,

$PR^2=PQ^2+QR^2\\\Rightarrow PR=\sqrt{PQ^2+QR^2}=\sqrt{24^2+24^2}=\sqrt{2\times24^2}=24\sqrt2$ m

If we move P towards Q 9m then PQ = 24-9 = 15 m

If the angle of elevation of the top of the building is $\phi$, then,

$\tan\phi=\frac{QR}{PQ}=\frac{24}{15}=1.6$

Hence the correct answers are,

(i) The height of the building logo from ground is (c) 8 √3 m.

(ii) The height of the building from ground is (c) 24 m.

(iii) The aerial distance of the point P from the top of the building is (b) 24√2 m.

(iv) In above case the angle of elevation ϕ of the top of building is given by (a) tan ϕ = 1.6.

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