Question 7·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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In right triangle $ABC$ , the right angle is at $C$ . Point $D$ lies between $A$ and $C$ on $\overline{AC}$ . Suppose that

\tan \angle BAC=\frac{5}{12}

and

\tan \angle DBC=\frac{3}{4}.

If $AD=66$ , what is the length of $BD$ ?

Your answer

Two tangent ratios in right triangles that share a leg point to one scale factor. Name the opposite and adjacent legs from each given angle, then use the segment between $A$ and $D$ to make an equation for Desmos. Watch the part-whole relationship: $AD$ is only part of $AC$ , and the requested $BD$ is a hypotenuse, not a leg.

Hints

Hint 1
For tangent, divide the opposite leg by the adjacent leg. Looking from $A$ , $BC$ is opposite and $AC$ is adjacent. Write their lengths using the same positive scale factor.
Hint 2
Triangle $BCD$ is right at $C$ because $D$ lies on $AC$ . From angle $DBC$ at $B$ , the opposite leg is $DC$ , not $AD$ . Which leg is adjacent?
Hint 3
The given $AD$ is a difference: $AC=AD+DC$ . Use the two tangent ratios to express $AC$ and $DC$ with one scale factor, then make an equation using $AD=66$ .

Step-by-step

Scale the tangent ratios
Step 1
Express the larger triangle's legs
At $A$ , tangent means opposite leg divided by adjacent leg. Those legs are $BC$ and $AC$ , so their lengths have the ratio $5:12$ . Let $x_1>0$ be their shared scale factor: $BC=5x_1,\qquad AC=12x_1$ .
Step 2
Express the smaller triangle's leg
Since $D$ lies on $AC$ , triangle $BCD$ is also right at $C$ . From angle $DBC$ , the opposite leg is $DC$ and the adjacent leg is $BC$ , so tangent gives: $\frac{DC}{BC}=\frac34$ . Substitute $BC=5x_1$ , then multiply: $DC=\frac34(5x_1)$ $DC=\frac{15}{4}x_1$ .
Step 3
Use the length between A and D
Because $D$ is between $A$ and $C$ , subtract $DC$ from $AC$ to get $AD$ : $AD=AC-DC$ . Substitute the lengths you found and the given $AD=66$ : $66=12x_1-\frac{15}{4}x_1$ . Setting $AC=66$ would leave out $DC$ .
Step 4
Find the scale factor
Type $12x_1-\frac{15}{4}x_1\sim66$ . The $\sim$ tells Desmos to solve for $x_1$ rather than graph it. Under PARAMETERS, Desmos shows $x_1=8$ ; that's the scale factor, not a side length.
Step 5
Find the shared leg
Since $BC=5x_1$ , add $5x_1$ on the next Desmos line. It prints $40$ , so $BC=40$ . Don't stop there: $BC$ is a leg, while the question asks for $BD$ .
Step 6
Use the smaller triangle's side ratio
In right triangle $BCD$ , the tangent ratio makes $DC:BC=3:4$ . The Pythagorean triple $3$ - $4$ - $5$ says its hypotenuse $BD$ has $5$ parts. Since $BC=40$ has $4$ parts, $BD=\frac54(40)=50$ . The length of $BD$ is $50$ . Grid in 50.

Question 7·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

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