Question

If the angles of elevation of the top of the candle from two coins distant ‘a’
cm and ‘b’ cm (a>b) from its base and in the same straight line from it are
30˚ and 60˚, then find the height of the candle

Answer 1

First of All

Let the case in form of a right ∆ABC & ∆ABD, both on same side of AB.

So ATQ

Let AB be the candle,

BD= b,

BC = a,

∠ADB = 60°

∠ACB = 30°

To find :

Height of candle, AB

Solution :

$\mathrm{In \: \triangle ABD}\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{tan \: D = \frac{AB}{BD}}\\ \implies \mathrm{tan \: 60 ° = \frac{AB}{b}}\\( \: \because tan \: 60° = \sqrt{3} \: ) \\ \implies \: \: \: \: \: \: \: \: \mathrm{\sqrt{3} = \frac{AB}{b} }\\ \implies \orange{\mathbb{AB = \sqrt{3}b}}--(a) \\ \\ Similarly, \: \mathrm{In \: \triangle ABC}\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{tan \: C= \frac{AB}{BC} } \\ \implies \mathrm{tan \: 30° = \frac{AB}{a} }\\ ( \: \because \mathrm{tan \: 30°}= \frac{1}{\sqrt{3}} \: )\\ \implies \: \: \: \: \: \: \: \mathrm{\frac{1}{ \sqrt{3} } = \frac{AB}{a} }\\ \: \: \:\implies \mathrm{\blue{AB= \frac{a}{ \sqrt{3} } } --(b)}$

Multiplying (a) with (b)

We get

$\mathrm{AB \: • \: AB= \sqrt{3}b \times \frac{a}{ {\sqrt{3} }} }\\ \implies \mathrm{{AB}^{2}= ab} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: \: \orange{\boxed{\green {\mathrm{AB = \sqrt{ab} \: \: \: ✓ } }}} \: \ \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

Answer:

$\sqrt{ab}$ cm is the required height of candle

Step-by-step explanation:

Explanation:

Given , there are two coins one at a distance of 'a' cm  from the candle and  the other is at 'b' cm from the candle .

Angles of elevation of the top of the candle from two coins are 30 °and 60°.

Let AB be the candle  whose height is 'h' cm and let   C and D  be the coins .

Therefore , ∠ACB = 30° and ∠ADB = 60 °

Step 1:

Now in triangle Δ ADC

tan 60° = $\frac{AB}{BD} = \frac{h}{b}$            

⇒$\sqrt{3} = \frac{h}{b}$ ⇒   $h = b\sqrt{3}$........(i)            (value of tan 60 ° is $\sqrt{3}$)

Similarly , In Δ ACB we have

tan30° = $\frac{AB}{BC} = \frac{h}{a}$

⇒$\frac{1}{\sqrt{3} } = \frac{h}{a}$   ⇒ $h = \frac{a}{\sqrt{3} }$ ........(ii)            (value of tan 30 ° is $\frac{1}{\sqrt{3} }$)

Step2:

Multiply  (i) and (ii) we get

$h^{2} = b\sqrt{3} .\frac{a}{\sqrt{3} }$

⇒$h^{2} = ab$  

⇒$h = \sqrt{ab}$ cm  = AB

Final answer :

Hence , height of the candle is $\sqrt{ab} cm$

#SPJ3