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Chapter 8: Q. 14 (page 535)

Solve the triangle

Short Answer

Expert verified

The solution is

$A = 125.09 °, B = 30.68 °, C = 24.23 °$

Step by step solution

01

Step1. Given information

The figure is

02

Step2. Formula used

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$

03

step3. Calculation

We have

$a = 8, b = 5, c = 4 W e k n o w, \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{5^{2} + 4^{2} - 8^{2}}{2.5.4} = \frac{25 + 16 - 64}{40} = \frac{- 23}{40} = - . 575 A = \cos^{- 1} (- . 575) = 125.09 ° A g a i n, \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{8^{2} + 4^{2} - 5^{2}}{2.8.4} = \frac{64 + 16 - 25}{64} = \frac{55}{64} = . 86 B = \cos^{- 1} (. 86) = 30.68 ° A g a i n, A + B + C = 180 ° 125.09 ° + 30.68 ° + C = 180 ° C = 180 ° - 155.09 = 24.23 °$

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Most popular questions from this chapter

Landscaping Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. The angle of elevation of the tree from one

position on a flat path from the tree is 30°, and from a second position 40 feet farther along this path it is 20°. What is the height of the tree?

A special case of the Law of Cosines is the Pythagorean theorem.

Area of an ASA Triangle If two angles and the included side are given, the third angle is easy to find. Use the Law of Sines to show that the area K of the triangle with side $a$ and angles A, B, and C is

$K = \frac{a^{2} \sin B \sin C}{2 \sin A}$

In the given problem solve the triangle using either the law of sines or law of cosines-

$a = 14, b = 7, A = 85 °$

Find an exact value of each expression, without using a calculator $\cos 62 - \sin 28$

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