3 years ago

HELP ASAPPPP !!!!!!!!!!!......

Mathematics

1 answer:

irina [24]3 years ago

8 0

Answer:

just use a calculator

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The vertices of a triangle ABC are A(7, 5), B(4, 2), and C(9, 2). What is measure of angle ABC? 30° 45° 56.31° 78.69°

GenaCL600 [577]

The measure of angle ABC is 45°

Explanation

Vertices of the triangle are: A(7, 5), B(4, 2), and C(9, 2)

According to the diagram below....

Length of the side BC (a) $=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5$

Length of the side AC (b) $= \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}$

Length of the side AB (c) $= \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}$

We need to find ∠ABC or ∠B . So using Cosine rule, we will get...

$cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree$

So, the measure of angle ABC is 45°

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