KnowledgeBoat Logo
|

Mathematics

In the given figure, ac = y units, BC = x units, then the height of DC is :

  1. x+y\sqrt{x + y}

  2. xy\sqrt{xy}

  3. y−x\sqrt{y - x}

  4. yx\sqrt{\dfrac{y}{x}}

In the given figure, ac = y units, BC = x units, then the height of DC is : Model Question Paper - 2, Concise Mathematics Solutions ICSE Class 10.

Trigonometric Identities

20 Likes

Answer

In △ ACD,

⇒ tan 30° = DCAC\dfrac{DC}{AC}

⇒ 13=DCy\dfrac{1}{\sqrt{3}} = \dfrac{DC}{y} …….(1)

In △ BCD,

⇒ tan 60° = DCBC\dfrac{DC}{BC}

⇒ 3=DCx\sqrt{3} = \dfrac{DC}{x} ……….(2)

Multiplying equation (1) and (2), we get :

⇒13×3=DCy×DCx⇒1=DC2xy⇒DC2=xy⇒DC=xy.\Rightarrow \dfrac{1}{\sqrt{3}} \times \sqrt{3} = \dfrac{DC}{y} \times \dfrac{DC}{x} \\[1em] \Rightarrow 1 = \dfrac{DC^2}{xy} \\[1em] \Rightarrow DC^2 = xy \\[1em] \Rightarrow DC = \sqrt{xy}.

Hence, Option 2 is the correct option.

Answered By

3 Likes

Related Questions