Note that
(4sqrt3)^2 = (2sqrt3)^2 + 6^2 [ 48 = 12 + 36] so the triangle is a right angled one with hypotenuse AB

So angle C = 90 degrees

sin A = 2 sqrt3 / 4 sqrt3 = 1.2 giving A = 30 degrees
and B = 90 - 30 = 60 degrees

A = 30, B = 60 and C = 90 degrees

5 0

3 years ago

Given: ΔPSQ, PS = SQ

horrorfan [7]

<h2>Answer:</h2>

To solve this problem we will use Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $a, \ b \ and \ c$ are the side lengths of the triangle and $s$ is the semiperimeter (half the perimeter of the triangle). We know that:

$Perimeter \ P=\triangle PSQ=PS+PQ+SQ: \\ \\ \triangle PSQ=P=50 \\ \\ Semiperimeter \ s: \\ \\ s=\frac{P}{2}=25$

Also:

$(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}$

Finally:

$A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}$

7 0

3 years ago

The figure is drawn on centimeter grid paper. Find the perimeter of the shaded figure to the nearest tenth.

nevsk [136]

I hope this helps you

8 0

3 years ago

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