Get It Right

Geometry Level 2

Given that triangle A B C ABC has no axes of symmetry and sin ⁡ ∠ A B C = cos ⁡ ∠ A C B \sin { \angle ABC } =\cos { \angle ACB } , find the value of sin ⁡ ∠ B A C + cos ⁡ ∠ B A C \sin { \angle BAC+\cos { \angle BAC } } .


The answer is 1.

Lee Care Gene by Lee Care Gene , 20, Malaysia

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1 solution

Marta Reece
May 7, 2017

If sin ⁡ ∠ A B C = cos ⁡ ∠ A C B \sin { \angle ABC } =\cos { \angle ACB } , that means that the two angles add up to 9 0 ∘ 90^\circ .

The remaining angle of the triangle, that is ∠ B A C \angle BAC , is also 9 0 ∘ 90^\circ .

Therefore sin ⁡ ∠ B A C + cos ⁡ ∠ B A C = sin ⁡ 9 0 ∘ + cos ⁡ 9 0 ∘ = 1 + 0 = 1 \sin { \angle BAC+\cos { \angle BAC } }=\sin90^\circ+\cos90^\circ=1+0=\boxed{1}

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