All categories

3 years ago

work out the total area of triangle ABC. calculate the area of AC. sides are:8.2cm and 13.5cm and angle 81 degrees

Mathematics

1 answer:

umka2103 [35]3 years ago

7 0

Answer:

54.67 cm²

Step-by-step explanation:

Given that the two sides of Δ ABC are 8.2 cm and 13.5 cm and the angle included by those sides is 81°.

Let, b = 8.2 cm and c = 13.5 cm and ∠ A = 81°

Hence, area of triangle ABC, Δ = $\frac{1}{2} bc \sin A$

⇒ Δ = $\frac{1}{2} bc \sin 81$

⇒ Δ = $\frac{8.2 \times 13.5 \times \sin 81}{2}$

⇒ Δ = 54.67 cm² (Answer)

You might be interested in

if leah is 6 years old then sue and John is 5 years older than leah and the total of the ages is 41 then how old is sue

lora16 [44]

6+5+x=41 11+x=41 11+-11+x=41+-11 x=30 Sue is 30 years old

7 0

2 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago

point p is located at (x, y). The point is reflected in the y-axis. Point P’s image is located at (__?__)

lesya [120]

Answer:

-x,y?

Step-by-step explanation:

if it's being reflected across the y-axis then that should go in to the negative side of the graph and the same if its being reflected across the x-axis, in which case it would be x,-y

3 0

3 years ago

Read 2 more answers

What was the first public statement in support of the creation of a Jewish state?

Ira Lisetskai [31]

The answer is D the Balfour Declaration

8 0

3 years ago

Definition of a line segment?

Nostrana [21]

Part of a line but has 2 end point, extends in no direction. example: @[email protected]

5 0

3 years ago