In triangle ABC, ∠C=90∘. Point D is the midpoint of BC, and AD=BC. Find sin∠BAD.
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Step-by-Step Solution
Step 1: Define variables and relationships
Let's assign a variable to the length of CD. Since D is the midpoint of BC, CD=BD. Therefore, BC=2CD. We are given that AD=BC, so AD=2x.
Step 2: Apply Pythagorean theorem in △ADC
In the right-angled triangle ADC, we can apply the Pythagorean theorem. We have AC2+CD2=AD2. Substituting the values we defined, we get AC2+x2=(2x)2. Solving for AC, we find AC=x3.
Step 3: Apply Pythagorean theorem in △ABC
Next, let's consider the larger right-angled triangle ABC. Using the Pythagorean theorem, AB2=AC2+BC2. Substituting the values we found, AB2=(x3)2+(2x)2. This simplifies to AB2=3x2+4x2=7x2, so AB=x7.
Step 4: Apply the Law of Cosines in △ABD
In triangle ABD, we can use the Law of Cosines to find cos∠ABD. The formula is AD2=AB2+BD2−2(AB)(BD)cos∠ABD. Substituting the lengths we found, we get (2x)2=(x7)2+x2−2(x7)(x)cos∠ABD. After simplifying, we find that cos∠ABD=72.
Step 5: Find sin∠ABD
Using the trigonometric identity sin2θ+cos2θ=1, we can find sin∠ABD. We have sin2∠ABD=1−(72)2=1−74=73. Taking the square root, sin∠ABD=73=721.
Step 6: Apply the Law of Sines in △ABD
Finally, we can use the Law of Sines in △ABD to find sin∠BAD. The formula is sin∠ABDAD=sin∠BADBD. Substituting the known values, we have 7212x=sin∠BADx. Solving for sin∠BAD, we get sin∠BAD=1421.