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Doss [256]
3 years ago
5

Jessica went shopping at a mall. She bought some dresses that were packed in a shopping bag. The shopping bag was in the shape o

f a trapezoid. The measure of ∠OPM is 10 less than three times the measure of ∠PMN. What is the measure of ∠OPM of the shopping bag?
Mathematics
1 answer:
olga2289 [7]3 years ago
6 0
Two angles of the adjacent side of a trapezoid should have a total angle of 180. In You can get this equation: 
∠OPM + ∠PMN= 180
∠OPM = 180- ∠PMN

∠OPM = 3(∠PMN) -10 
 180- ∠PMN= 3∠PMN -10 
 180+10= 3∠PMN + ∠PMN
 4  ∠PMN= 190
 ∠PMN= 47.5

∠OPM + ∠PMN= 180
∠OPM + 47.5= 180
∠OPM = 180-47.5= 132.5
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\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of \triangle ACD and the hypotenuse of \triangle ADB.

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is \angle CAD. The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}

Now use this value in the Law of Sines to find AD:

\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}

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AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}

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Since AD is the hypotenuse, it must represent 2x in this ratio and since AB is the side opposite to the 30 degree angle, it must represent x in this ratio (Derive from basic trig for a right triangle and \sin 30^{\circ}=\frac{1}{2}).

Therefore, AB must be exactly half of AD:

AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24

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