Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 11 - Review Exercises: 13

Answer

The surveyor is around $\approx 42.7 $ ft from the base of the building.

Work Step by Step

Looking at Figure #1 in the attachment: $A =$ Position of surveyor $B = $ Base of the building $C = $ Point horizontally across from the surveyor to the building $D = $ Top of the building In total, there are two right angled triangles present, $\triangle ABC$ and $\triangle ADC$. 1. Find the angles in all of the triangles $\triangle ABC$: $\angle C = 90˚$ (Properties of a right angled triangle) $\angle BAC = 16˚$ (Given by the question) $\angle B = [180 - (90 + 16)]$ (Angles in a triangle must add up to $180˚$) $\angle B = [180 - 106]$ $\angle B = 74˚$ $\triangle ADC$: $\angle C = 90˚$ (Properties of a right angled triangle) $\angle DAC = 43˚$ (Given by the question) $\angle D = [180 - (43 + 90)]$ (Angles in a triangle must add up to $180˚$) $\angle D = [180 - 133]$ $\angle D = 47˚$ 2. Verify that $\angle A$, $\angle B$ and $\angle D$ adds to $180˚$ $\angle A + \angle B+\angle D = 180˚$ $(43 + 16) + (74) + (47) = 180˚$ $59 + 74 + 47 = 180˚$ $180˚ = 180˚$ 3. Solve for $x$ using the sine law (Use a calculator) $\frac{50}{sin59} = \frac{x}{sin47}$ $x = \frac{50sin47}{sin59}$ $x = \frac{36.5676...}{0.8571...}$ $x \approx 42.7$ ft
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