I have problem with solving this task.
I know that the answer might be A. But only with calculator by calculating the angles. can someone explain me or give me a hint to solve it.
cos^-1 (5/13) = 67.38° cos (67.38°x2) = -0.704 = 119/169
$\endgroup$ 14 Answers
$\begingroup$Well, you know all the side lengths, and you know that $|BD| = 10$.
So this gives you two equal triangles, $ADB$ and $CDB$. You know all the side lengths. Split them in half to make right-angle triangles. Now you've split parallelogram $ADBC$ into quarters. All of these right-angle triangles have the same measurements. Pick any one of them. You know two of the side lengths ($13,5$). Use the Pythagorean Theorem to find the third length ($13^2-5^2 = 169-25=144, \sqrt{144} = 12$). You know one of the angles ($90$ deg.). Use your favorite trigonometric rule to figure out the angle that you need (this will be an angle on $D$ or $B$), then just add two of those together and you're done.
If my instructions seem unclear: draw a diagram of the situation and follow step by step with a pen.
$\endgroup$ $\begingroup$An equilateral quadrilateral is a rhombus. The diagonals of a rhombus are perpendicular bisectors of each other. Thus, they divide the rhombus into four congruent right triangles. If the diagonals intersect at point $E$, then
$$BE = \frac{1}{2}BD = 5~\text{cm}$$
Applying the Pythagorean Theorem to $\triangle ABE$ yields $AE = 12~\text{cm}$. Since $AC = 2AE$, $AC = 24~\text{cm}$.
Apply the Law of Cosines to $\triangle ABC$. Since you know $AC$, $AB$, and $BC$, you can solve for $\cos(\angle ABC)$.
$\endgroup$ $\begingroup$In the rhombus diagonals bisect orthogonally.
$$ \cos \angle DBA= c = \frac {5}{13}$$
Since $ \angle ABC = 2 \angle ABD,$ cosine double angle
$$ \cos \angle ABC = 2 c^2-1 = \frac {-111}{169}. $$
$\endgroup$ $\begingroup$(A) is correct
$\cos ABC= cos2\theta=2(\cos \theta)^2-1=\frac{2\cdot 5^2}{13^2}-1=-\frac{119}{169}$
