In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If BAC = 45^ and ABC =, then AD/BE equals Option 1: √2 Option 2: 1 Option 3: ( + )/√2 Option 4: √2

Answer: √2. Solution: We have triangle ABC where BAC = 45°, ABC = θ, AD is altitude from vertex A to side BC, BE is altitude from vertex B to side AC, and we need to find AD/BE. Both altitudes help us calculate the same area of triangle ABC, just using different base-height pairs. Area of triangle ABC can be expressed in two ways: Using base BC and height AD: Area = 1/2 x BC x AD Using base AC and height BE: Area = 1/2 x AC x BE Since both expressions equal the same area: 1/2 x BC x AD = 1/2 x AC x BE BC x AD = AC x BE Therefore: AD/BE = AC/BC Now we need to find the ratio AC/BC using our given angles. Law of Sines states: In any triangle, the ratio of a side to the sine of its opposite angle is constant. For triangle ABC: AC/(( ABC)) = BC/(( BAC)) AC/((θ)) = BC/((45°)) Since (45°) = 1/√2: AC/((θ)) = BC/(1/√2) AC/BC = ((θ))/((45°)) = ((θ))/(1/√2) = √2 (θ) Since AD/BE = AC/BC, we have: AD/BE = √2 (θ)

Geometry

Triangles

Medium

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^\circ$ and $\angle ABC = \theta$ , then $\frac{AD}{BE}$ equals

$\sqrt{2} \cos \theta$

$\frac{(\sin \theta + \cos \theta)}{\sqrt{2}}$

$\sqrt{2} \sin \theta$

✅ Correct Option: 4

We have triangle ABC where $\angle BAC = 45°$ , $\angle ABC = θ$ , AD is altitude from vertex A to side BC, BE is altitude from vertex B to side AC, and we need to find $\tfrac{AD}{BE}$ .

Both altitudes help us calculate the same area of triangle ABC, just using different base-height pairs.

Area of triangle ABC can be expressed in two ways:

Using base BC and height AD: Area $= \frac{1}{2} \times BC \times AD$

Using base AC and height BE: Area $= \frac{1}{2} \times AC \times BE$

Since both expressions equal the same area:

$\frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE$

$BC \times AD = AC \times BE$

Therefore: $\frac{AD}{BE} = \frac{AC}{BC}$

Now we need to find the ratio $\tfrac{AC}{BC}$ using our given angles.

Law of Sines states: In any triangle, the ratio of a side to the sine of its opposite angle is constant.

For triangle ABC:

$\frac{AC}{\sin(\angle ABC)} = \frac{BC}{\sin(\angle BAC)}$

$\frac{AC}{\sin(θ)} = \frac{BC}{\sin(45°)}$

Since $\sin(45°) = \frac{1}{\sqrt{2}}$ :

$\frac{AC}{\sin(θ)} = \frac{BC}{\frac{1}{\sqrt{2}}}$

$\frac{AC}{BC} = \frac{\sin(θ)}{\sin(45°)}$

$= \frac{\sin(θ)}{\frac{1}{\sqrt{2}}}$

$= \sqrt{2} \sin(θ)$

Since $\frac{AD}{BE} = \frac{AC}{BC}$ , we have:

$\frac{AD}{BE} = \sqrt{2} \sin(θ)$

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In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC=45∘\angle BAC = 45^\circ∠BAC=45∘ and ∠ABC=θ\angle ABC = \theta∠ABC=θ, then ADBE\frac{AD}{BE}BEAD​ equals

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In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^\circ$ and $\angle ABC = \theta$ , then $\frac{AD}{BE}$ equals