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3 years ago

How to solve for area of triangle where A=67 degrees, b=11yd, and c=24yd

Mathematics

1 answer:

skad [1K]3 years ago

4 0

Use the formula of an area of a tringle:

$A=\dfrac{1}{2}bc\sin A$

We have b = 11 yd, c = 24 yd and m∠A = 67°

$A=\dfrac{1}{2}(11)(24)\sin67^o$

$\sin67^o\approx0.9205\\\A=(11)(12)(0.9205)=121.506\ yd^2$

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0.6(4-2x)=20.5-(3x+10)

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The solution of 0.6(4-2x)=20.5-(3x+10) is x = 4.5

<u>Solution:</u>

Given, expression is 0.6 ( 4 – 2x ) = 20.5 – ( 3x + 10 )

We have to solve the above given equation for x.

Now, take the equation

$\rightarrow 0.6(4-2 x)=20.5-(3 x+10)$

We have to use BODMAS rule. BODMAS means sequence of operations which is brackets, order, division, multiplication, addition and then subtraction

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$\rightarrow 0.6 \times 4-0.6 \times 2 x=20.5-3 x-10$

Now multiplication is given preference

$\rightarrow 2.4-1.2 x=10.5-3 x$

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Hence, the value of x is 4.5

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