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elena-s [515]
4 years ago
7

PLEASEEE HELP ME ASAP!!

Mathematics
1 answer:
denpristay [2]4 years ago
7 0

Answer:

Hence,

The measure of ∠A is 60.065°

\m\angle A= 60.065

Step-by-step explanation:

Given:

ΔABC is a Right Angle Triangle at ∠ B = 90°

BC = Opposite side to ∠A = 13 unit

AC = Hypotenuse  = 15 unit  

To Find:

m∠A = ?  

Solution:

In Right Angle Triangle ABC ,Sine Identity,

\sin A= \dfrac{\textrm{side opposite to angle A}}{Hypotenuse}\\

Substituting the values we get

\sin A= \dfrac{BC}{AC}=\dfrac{13}{15}=0.8666\\\angle A=\sin^{-1}(0.8666)\\m\angle A= 60.065\°

Hence,

The measure of ∠A is 60.065°

\m\angle A= 60.065

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5 0
4 years ago
What is the area of ABC?
Salsk061 [2.6K]

Answer:

Area of triangle ABC = 10

Option C is correct.

Step-by-step explanation:

We need to find area of triangle ABC

We are given A(2,1) B(4,7) and C(6,3)

The formula used to find Area \ of \ triangle=\frac{1}{2}base\times height ABC is:

In the given triangle base= AC and height = BC

First we need to find the distance between A and C and B and C

The formula used to find distance between AC Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have x_1=2, y_1=1, x_2=6, y_2=3

Putting values and finding distance

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-2)^2+(3-1)^2}\\Distance=\sqrt{(4)^2+(2)^2}\\Distance=\sqrt{16+4}\\Distance=\sqrt{20}

find distance between BC Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have x_1=4, y_1=7, x_2=6, y_2=3

Putting values and finding distance

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-4)^2+(3-7)^2}\\Distance=\sqrt{(2)^2+(4)^2}\\Distance=\sqrt{4+16}\\Distance=\sqrt{20}

So, We have height = \sqrt{20} and base = \sqrt{20}

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So, Area of triangle ABC = 10

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