Question

The angle of elevation of the top of a
building from a point on the ground,
which is 30 m away from the foot of
the building, is 30°. The height of the
building is:
(A) 10 m
(B) 30/03 m
(C) 03/10 m
(D) 30 m​

Answer 1

given options are incorrect .

correct options are :-

1. 10 m
2. 30/√3 m
3. √3/30 m
4. 30 m

option (2) is correct . 30/√3 m

we have :-

• angle of elevation on the ground from top of the building = 30⁰
• the distance from foot of ground = 30m

To find :-

★ the height of the building .

Solution :-

the case in which from the top of building to the ground all these makes a right angle traingle .

so,

in ∆ABC

let AB be the height of building = h metre

$\sf \: as \: we \: know \: that \: \: \: \dfrac{perpendicular}{base} = tan \theta \\ \\ \\ \sf \dfrac{AB}{BC} = tan \theta \: \\ \\ \bf \: so \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \dfrac{h}{30} = tan 30 \degree \\ \\ \sf the\: value\: of \: \bf {tan 30\degree\: is \dfrac{1}{ \sqrt{3} } } \: \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{h}{30} = \frac{1}{ \sqrt{3} } \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h \sqrt{3} = 30 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = \dfrac{30}{ \sqrt{3} } \: m \:$

therefore height of building = $\sf<strong> </strong> h = \dfrac{30}{ \sqrt{3} } \: m$

more concepts :-

$\sf \red{sin \theta = \dfrac{perpendicular}{hypotenuse}} \\ \\ \\\sf \blue{cos \theta = \dfrac{base}{hypotenuse} }\\ \\ \\\sf\pink{ tan \theta = \dfrac{perpendicular}{base}} \\ \\ \\\sf \red{cot \theta = \dfrac{1}{tan\theta}=\dfrac{base}{perpendicular}} \\ \\ \\\sf\blue{ cosec \theta = \dfrac{1}{sin\theta} = \dfrac{hypotenuse}{perpendicular}}\\ \\ \\\sf\green{ sec \theta = \dfrac{1}{cos\theta}= \dfrac{hypotenuse}{base}}$

Answer 2

Answer:

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