Question

Difficulty: HardRight Triangle Trigonometry (SOHCAHTOA)

In right triangle XYZXYZ, the right angle is at vertex YY, and line segment YWYW is an altitude perpendicular to hypotenuse XZXZ at point WW. If the length of side XYXY is 1515 units and cos(X)=45\cos(X) = \frac{4}{5}, what is the length of line segment ZWZW?

  1. A
    99
  2. B
    1212
  3. 274\frac{27}{4}Answer
  4. D
    454\frac{45}{4}
  5. E
    1616

Answer

The length of line segment ZWZW is 274\frac{27}{4} units.
In right triangle XYZXYZ, cos(X)=XYXZ\cos(X) = \frac{XY}{XZ}. Given XY=15XY = 15 and cos(X)=45\cos(X) = \frac{4}{5}, we solve for hypotenuse XZ=754XZ = \frac{75}{4}. In right triangle XYWXYW, cos(X)=XWXY=XW15\cos(X) = \frac{XW}{XY} = \frac{XW}{15}, yielding XW=12XW = 12. Subtracting XWXW from total hypotenuse XZXZ yields ZW=75412=274ZW = \frac{75}{4} - 12 = \frac{27}{4}.

Step-by-Step Solution

1
Find the length of hypotenuse XZXZ using cos(X)\cos(X) in XYZ\triangle XYZ.
XZ=754XZ = \frac{75}{4}
In XYZ\triangle XYZ, cos(X)=adjacenthypotenuse=XYXZ\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{XZ}. Substituting cos(X)=45\cos(X) = \frac{4}{5} and XY=15XY = 15 gives 45=15XZ\frac{4}{5} = \frac{15}{XZ}, so XZ=15×54=754XZ = \frac{15 \times 5}{4} = \frac{75}{4}.
2
Find the length of segment XWXW using cos(X)\cos(X) in right triangle XYWXYW.
XW=12XW = 12
In right triangle XYWXYW (with right angle at WW), cos(X)=adjacenthypotenuse=XWXY\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XW}{XY}. Substituting cos(X)=45\cos(X) = \frac{4}{5} and XY=15XY = 15 gives 45=XW15\frac{4}{5} = \frac{XW}{15}, so XW=12XW = 12.
3
Calculate segment ZWZW by subtracting XWXW from total hypotenuse XZXZ.
ZW=274ZW = \frac{27}{4}
Since point WW lies on segment XZXZ, ZW=XZXW=75412=754484=274ZW = XZ - XW = \frac{75}{4} - 12 = \frac{75}{4} - \frac{48}{4} = \frac{27}{4}.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) in Nested Right Triangles
Rate this question