NSEJS Practise problem 1

Geometry Level 1

A triangle has sides of length sin ⁡ ( x ) , cos ⁡ ( x ) \sin( x), \cos (x) and 1 + sin ⁡ ( x ) cos ⁡ ( x ) \sqrt{1+\sin (x )\cos( x)} , with x x an acute angle.

Find the largest angle of the triangle in degrees.


The answer is 120.

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Parth Lohomi by Parth Lohomi , 20, India

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1 solution

Using the Cosine Rule, we get: c o s ( C ) = a 2 + b 2 − c 2 2 a b = s i n 2 ( x ) + c o s 2 ( x ) − ( 1 + s i n ( x ) c o s ( x ) ) 2 s i n ( x ) c o s ( x ) \displaystyle cos(C) = \frac{a^2+ b^2 - c^2}{2ab} = \frac{sin^{2}(x) + cos^{2}(x) -(1+sin(x)cos(x))}{2sin(x)cos(x)}

Simplifying this gives:

c o s ( C ) = 1 − 1 − s i n ( x ) c o s ( x ) 2 s i n ( x ) c o s ( x ) = − 1 2 \displaystyle cos(C) = \frac{1-1-sin(x)cos(x)}{2sin(x)cos(x)} = \frac{-1}{2}

This implies that C = 12 0 ∘ \displaystyle C = 120^{\circ}

Note that we are done, as the remaining angles must be less than 120 ∘ ^{\circ} . (since sum of the angles is 180 ∘ ^{\circ} , and the remaining angles are acute)

Hence, that largest angle is 12 0 ∘ \displaystyle \boxed{120^{\circ}}

17 Helpful 1 Interesting 1 Brilliant 2 Confused

Moderator note:

Good approach using the cosine rule to solve for the triangle.

Addendum: C C is the largest angle since c = 1 + sin ⁡ x cos ⁡ x ≥ 1 ≥ sin ⁡ x , cos ⁡ x c = \sqrt{1+\sin x \cos x} \geq 1 \geq \sin x, \cos x is the largest side.

Jake Lai - 5 years, 11 months ago

Same Method

Kushagra Sahni - 5 years, 11 months ago

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