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a parallelogram has sides of lengths 4 and 9, and one angle is 41°. wha…

Question

a parallelogram has sides of lengths 4 and 9, and one angle is 41°. what is the length of the smaller diagonal? length = units what is the length of the longer diagonal? length = units enter your answer as a decimal; your answer should be accurate to 3 decimal places. question help: message instructor

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). In a parallelogram, the smaller diagonal is opposite the smaller - angle. Let \(a = 4\), \(b = 9\), and \(C=41^{\circ}\).

Step2: Calculate the length of the smaller diagonal

Substitute \(a = 4\), \(b = 9\), and \(C = 41^{\circ}\) into the Law of Cosines formula.
\(\cos(41^{\circ})\approx0.7547\)
\(d_{1}^{2}=4^{2}+9^{2}-2\times4\times9\times\cos(41^{\circ})\)
\(d_{1}^{2}=16 + 81-72\times0.7547\)
\(d_{1}^{2}=97-54.3384\)
\(d_{1}^{2}=42.6616\)
\(d_{1}=\sqrt{42.6616}\approx6.532\)

Step3: Calculate the length of the larger diagonal

The larger - angle in the parallelogram is \(180^{\circ}-41^{\circ}=139^{\circ}\), and \(\cos(139^{\circ})\approx - 0.7547\)
Using the Law of Cosines \(d_{2}^{2}=4^{2}+9^{2}-2\times4\times9\times\cos(139^{\circ})\)
\(d_{2}^{2}=16 + 81-72\times(-0.7547)\)
\(d_{2}^{2}=97 + 54.3384\)
\(d_{2}^{2}=151.3384\)
\(d_{2}=\sqrt{151.3384}\approx12.302\)

Answer:

The length of the smaller diagonal is \(6.532\) units.
The length of the longer diagonal is \(12.302\) units.