5. In the diagram shown below, BC is drawn tangent

Mathematics

1 answer:

Tatiana [17]2 years ago

6 0

9514 1404 393

Answer:

(2) 72°

Step-by-step explanation:

In this geometry, the angle at the tangent is half the measure of the intercepted arc.

∠CBD = (arc BD)/2 = 144°/2

∠CBD = 72°

<em>Additional comment</em>

Consider a point X anywhere on long arc BD. The inscribed angle at X will have half the measure of short arc BD, so will be 144°/2 = 72°. This is true regardless of the position of X on long arc BD. Now, consider that X might be arbitrarily close to point B. The angle at X is still 72°.

As X approaches B, the chord XB approaches a tangent to the circle at B. Effectively, this tangent geometry is a limiting case of inscribed angle geometry.

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From a window 20 feet above the ground, the angle of elevation to the top of a building across

Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

$\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},$

and from the right-angled triangle WAB, we have'

$\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.$