Expert: Josh - 11/5/2006

Question
Dear Josh,
My problem involves two composite triangles, PQR (smaller) and PQZ (larger), with the following dimensions: side PQ = 25, side PZ = 66, side QR = 25, angle Z = 10.5 degrees. How would I solve for angle QPR/QPZ and angle QRZ?

Answer
Dear Maya,

My drawing looks like this:
..........Q
.
.
.
.
P....................R...................Z

The cosine law states that for a triangle with sides a,b, and c and angle "q" opposite to side c,
c^2 = a^2 + b^2 - 2*a*b·cos(q).

Thus, if we let a=|PZ|, b=|QZ| (which is unknown), c=|PQ| and q=^RZQ, we have
|PQ|^2 = |PZ|^2 + |QZ|^2 - 2*|PZ|*|QZ|*cos(q).

Substituting the known values |PQ|=25, |QR|=25, |PZ|=66, ^RZQ=10.5 [cos(10.5)=0.983254907563955...] into the equation, we obtain

25^2 = 66^2 + |QZ|^2 - 2*66*0.983254907563955*|QZ|
625 = 4356 + |QZ|^2 - 129.789647798442*|QZ|
|QZ|^2 - 129.789647798442*|QZ| + 3731 = 0

A quadratic equation in terms of length |QZ| can be solved in various ways. You can apply the quadratic formula, where A=1, B=-129.789647798442, C=3731 to obtain solutions |QZ|=[-B+sqrt(B^2-4AC)]/(2A),[-B-sqrt(B^2-4AC)]/(2A)

Computation gives two possible solutions:
|QZ|=86.81144245..., or |QZ|=42.97820534...

From the diagram, it is impossible for side |QZ| to be longer than |PZ|=66. We discard the first solution. Only the second solution makes sense. We retain |QZ|=42.97820534 as the approximate solution (accurate to 8 decimal places).

To find angle ^QRZ, we can let "w"=^QRZ and apply the sine rule, where |QZ|/sin(^QRZ)=|QR|/sin(^RZQ).

This is same as 42.97820534/sin(w)=25/sin(10.5) after substituting the known values. Rearrange equation to sin(w)=42.97820534*(sin(10.5)/25), taking the inverse sine function of (42.97820534*sin(10.5)/25), we get 18.257386... degrees. From picture, w=^QRZ is clearly an obtuse (not acute) angle. i.e., it is between 90 and 180 degrees. Recall that sin(180-w) also has the same value. This is because all angles in the first quadrant (betw. 0 and 90 deg) and second quadrant (betw. 90 and 180 deg) subtend a positive ratio.

So, in fact, ^QRZ = 180-18.257386 degrees.

Now, ^QRP=^QPR since QRP is an isosceles triangle.
^PRZ = ^PRQ + ^QRZ
^PRQ = 180 - ^QRZ = 18.2573686 degrees approximately.

If I made a mistake, I hope you would be able to correct it following a similar approach.