One side and two angles

Geometry Level 3

Find the area of the triangle ( \big( in unit 2 ) \text{unit}^2\big) whose two angles are 3 0 ∘ 30^{\circ} and 4 5 ∘ 45^{\circ} and the side included between them has length 3 + 1 \sqrt3 + 1 units.

1 3 ( 3 + 1 ) \dfrac{1}{3} \big( \sqrt3 + 1 \big) 1 2 ( 2 + 1 ) \dfrac{1}{2} \big( \sqrt2 + 1 \big) 1 2 ( 3 + 1 ) \dfrac{1}{2} \big( \sqrt3 + 1 \big) 1 2 ( 3 − 1 ) \dfrac{1}{2} \big( \sqrt3 - 1 \big)

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Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Chew-Seong Cheong
Dec 26, 2016

Referring to the figure above. Let the height of the triangle be h h , then we note that:

h 3 + 1 − h = tan ⁡ 3 0 ∘ h 3 + 1 − h = 1 3 3 h = 3 + 1 − h 3 h + h = 3 + 1 h = 3 + 1 3 + 1 = 1 \begin{aligned} \frac h{\sqrt 3+1-h} & = \tan 30^\circ \\ \frac h{\sqrt 3+1-h} & = \frac 1{\sqrt 3} \\ \sqrt 3 h & = \sqrt 3+1-h \\ \sqrt 3 h + h & = \sqrt 3+1 \\ h & = \frac {\sqrt 3+1}{\sqrt 3+1} = 1 \end{aligned}

Therefore, the area of the triangle = 1 2 h ( 3 + 1 ) = 1 2 ( 3 + 1 ) = \dfrac 12 h (\sqrt 3+1) = \boxed{\dfrac 12 (\sqrt 3+1)}

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B y By l a w law o f of s i n e s , sines,

x s i n 45 \frac{x}{sin 45} = = 3 + 1 s i n 105 \frac{\sqrt{3}+1}{sin 105}

x = 2 x=2

A r e a = Area= 1 2 \frac{1}{2} ( 2 ) ( 3 + 1 ) ( s i n 30 ) (2)(\sqrt{3}+1)(sin30) = = ( 3 + 1 ) ( 1 2 (\sqrt{3}+1)(\frac{1}{2} )

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