Problem:
Right triangle ACD with right angle at C is constructed outwards on the hypotenuse AC of isosceles right triangle ABC with leg length 1 , as shown, so that the two triangles have equal perimeters. What is sin(2∠BAD) ?
Answer Choices:
A. 31
B. 22
C. 43
D. 97
E. 23 Solution:
Firstly, note by the Pythagorean Theorem in △ABC that AC=2. Now, the equal perimeter condition means that BC+BA=2=CD+DA, since side AC is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in △ACD, gives AC2+CD2=(2)2+(2−DA)2=DA2. Hence 2+4−4DA+DA2=DA2, so DA=23, and thus CD=21.
Next, since ∠BAC=45∘,sin(∠BAC)=cos(∠BAC)=21. Using the lengths found above, sin(∠CAD)=(23)(21)=31, and cos(∠CAD)=(23)2=322.
Thus, by the addition formulae for sin and cos, we have
cos(∠BAD)=cos(∠BAC+∠CAD) =cos(∠BAC)cos(∠CAD)−sin(∠BAC)sin(∠CAD) =21⋅322−21⋅31 =3222−1
Hence, by the double angle formula for sin,sin(2∠BAD)=2sin(∠BAD)cos(∠BAD)=182(8−1)=(D)97.
OR
We use the Pythagorean Theorem, as in Solution 1, to find AD=23 and CD=21. Now notice that the angle between CD and the vertical (i.e. the y-axis) is 45∘ - to see this, drop a perpendicular from D to BA which meets BA at E, and use the fact that the angle sum of quadrilateral CBED must be 360∘. Anyway, this implies that the line CD has slope 1, so since C is the point (0,1) and the length of CD is 21,D has coordinates ⎝⎜⎜⎜⎛0+2(21),1+2(21)⎠⎟⎟⎟⎞=(221,1+221).
Thus we have the lengths DE=1+221 (it is just the y-coordinate) and AE=1−221. By simple trigonometry in △DAE, we now find
sin(∠BAD)=(23)(1+221)=3(2+21)=3222+1
and
cos(∠BAD)=(23)(1−221)=3(2−21)=3222−1
just as before. We can then use the double angle formula (as in Solution 1) to deduce sin(2∠BAD)=(D)97.
