Question

If two angles of a triangle are 30° and 45° and the included side is 2.732 m, find the area of triangle. Use sqrt(3) = 1.732​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, two angles of a triangle are 30° and 45° and the included side is 2.732 m

Let assume that the required triangle be ABC and let AD is perpendicular drawn from A to BC intersecting BC at D.

Let assume that AD = h m, ∠B = 45° and ∠C = 30°.

Now, to find the area of triangle ABC, we have to first find the height of triangle ABC.

So, Consider right-angled triangle ABD

$\rm \: tan45 \degree \: = \: \dfrac{AD}{BD}$

$\rm \: 1 \: = \: \dfrac{h}{BD}$

$\rm\implies \:BD \: = \: h \: - - - (1)$

Now, Consider right-angled triangle ADC

$\rm \: tan30 \degree \: = \: \dfrac{AD}{DC}$

$\rm \: \dfrac{1}{ \sqrt{3} } \: = \: \dfrac{h}{DC}$

$\rm\implies \:DC \: = \: \sqrt{3} \: h \: - - - (2)$

As it is further given that

$\rm \: BC = 2.732$

$\rm \: BD + DC = 2.732$

$\rm \: h + \sqrt{3} \: h = 2.732$

$\rm \: h(1 + \sqrt{3}) = 2.732$

$\rm \: h(1 + 1.732) = 2.732$

$\rm \: 2.732h = 2.732$

$\rm\implies \:h \: = \: 1 \: m$

Now, In triangle ABC, we have

$\rm \: BC \: = \: 2.732 \: m$

$\rm \: AD \: = \: 1 \: m$

So,

$\rm \: Area_{(\triangle\:ABC)} \: = \: \dfrac{1}{2} \: \times \: BC \: \times \: AD \:$

$\rm \: Area_{(\triangle\:ABC)} \: = \: \dfrac{1}{2} \: \times \: 2.732 \: \times \: 1 \:$

$\rm\implies \: \:\boxed{\sf{ \: \: \rm \: Area_{(\triangle\:ABC)} \: = \:1.366 \: \: }}$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\large\mathbb \red{\fcolorbox{red}{lime}{ ANSWER\ }}$