First slide

Applications of trigonometry

Question

If each side of length a of an equilateral triangle subtends an angle of $60^{0}$ at the top of a tower h meter high situated at the center of the triangle, then

Moderate

A

$3 a^{2} = 2 h^{2}$
B

$2 a^{2} = 3 h^{2}$
C

$a^{2} = 3 h^{2}$
D

$3 a^{2} = h^{2}$

Solution

          Let O be the center of the equilateral triangle ABC and OP be the tower of height h. Then each of the triangle PAB, PBC and PAC are equilateral and thus PA=PB=AB=a .
In triangle ABC,OA is the bisector of the angle A,
   $S o \frac{O A}{a / 2} = \sec 30^{0} \Rightarrow O A = \frac{a}{2} . \frac{2}{\sqrt{3}} = \frac{a}{\sqrt{3}}$
Now from right angle triangle POA
   $\begin{array}{l} P A^{2} = O P^{2} + O A^{2} \Rightarrow a^{2} = h^{2} + \frac{a^{2}}{3} \\ \Rightarrow \frac{2 a^{2}}{3} = h^{2} \Rightarrow 2 a^{2} = 3 h^{2} \end{array}$

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