# A 12 cm segment AB is divided by points M and N into three equal parts.

**A 12 cm segment AB is divided by points M and N into three equal parts. The projection of the segment MN onto the ray AC is 2 cm. Find the cosine of the angle BAC.**

Take an angle of value α with apex at point A. On one of the sides of this angle, take point B, and inside the resulting segment AB, mark points M and N so that they divide this segment into three equal parts:

| AM | = | MN | = | NB | = 1/3 * | AB |;

Draw perpendiculars MK, NL and BD from points M, N and B to the other side of the corner, on the AC line. By the condition of the problem:

| AB | = 12 (cm);

and the projection KL of the segment MN is 2 cm:

| KL | = 2 (cm);

Note that the points K, L and D lie on the ray AC. The task requires to calculate the cosine of the angle BAC or cosα.

Similarity of triangles

Consider three triangles – MKA, NLA and BDA. These triangles are:

are rectangular because MK ⊥ AK, NL ⊥ AL and BD ⊥ AD, and ∠MKA = ∠NLA = ∠BDA = 90 °;

have the same common angle α = ∠MAK = ∠NAL = ∠BAD;

have equal third angles ∠KMA = ∠LNA = ∠DBA = 90 ° – α.

Thus, since all three angles of these triangles are equal, then they are similar, and, therefore:

| BA | / | DA | = | NA | / | LA | = | MA | / | KA |;

Calculation of cosα

As you know, the ratio of the leg to the hypotenuse is equal to the cosine of the angle between them. Accordingly, for triangles MKA and NLA we get:

| LA | / | NA | = cosα;

| KA | / | MA | = cosα;

Further we have:

| LA | = | NA | * cosα;

| KA | = | MA | * cosα;

Subtracting the second from the first equality, we get:

| LA | – | KA | = (| NA | – | MA |) * cosα;

or

| KL | = | MN | * cosα;

By the condition of the problem:

| MN | = 1/3 * | AB | = 1/3 * 12 = 4 (cm);

| KL | = 2 (cm);

Substituting these values, we find the cosine of the required angle:

cosα = | KL | / | MN | = 2/4 = 1/2;

Answer: The cosine of the angle BAC is 1/2.