N the diagram, point O is the center of the circle and mADB = 43°. If mAOB = mBOC, what is mBDC?

Mathematics

1 answer:

Sonja [21]2 years ago

3 0

The answer is 43°
Angle ADB is an inscribed angle which means arc AB has an angle twice that of angle ADB. The angle of the arc would be the same as that of the central angle AOB. So, mAOB = 86°. And since, mAOB = mBOC, then mBOC = 86° and arc BC has a measure of 86° as well. Angle BDC intercepts the arc BC which means half of the angle of arc BC is mBDC. So, mBDC = 43.

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Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places

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The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

$\rm 5 \sin \theta = 4\\\\\theta = 0.927 , 2.214$

Then by the integration, we have

$\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\$

$\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\$