Math

Question(a) Find ADAD given tan50=AD12\tan 50^{\circ} = \frac{AD}{12}, where AD=14.30AD = 14.30. (b) Calculate angle BACBAC and verify it rounds to 40.4240.42^{\circ}. (c) Determine the area of quadrilateral ABCDABCD. (d) Find the shortest distance from BB to line ACAC.

Studdy Solution

STEP 1

Assumptions1. The given problem is a trigonometry problem involving a quadrilateral ABC. . The lengths and angles provided are accurate.
3. We are assuming the quadrilateral ABC to be a combination of two triangles, ABD and ABC.

STEP 2

First, we need to calculate the length of AD. We can use the tangent of the angle and the opposite side to find the adjacent side.
AD=12×tan50AD =12 \times \tan50^{\circ}

STEP 3

Calculate the length of AD.
AD=12×tan50=14.30cmAD =12 \times \tan50^{\circ} =14.30 \, cm

STEP 4

Next, we need to calculate the angle BAC. We can use the sine rule to find the angle.
sinBAC=811\sin BAC = \frac{8}{11}

STEP 5

To find the angle BAC, we need to take the inverse sine of the ratio.
BAC=sin1(811)BAC = \sin^{-1} \left(\frac{8}{11}\right)

STEP 6

Calculate the angle BAC.
BAC=sin1(811)=40.42BAC = \sin^{-1} \left(\frac{8}{11}\right) =40.42^{\circ}

STEP 7

Now, we need to calculate the area of the quadrilateral ABC. We can do this by calculating the area of the two triangles ABD and ABC separately and then adding them together.
AreaABCD=AreaAB+AreaABCArea_{ABCD} = Area_{AB} + Area_{ABC}

STEP 8

The area of a triangle is given by the formula 12×base×height\frac{1}{2} \times base \times height. For triangle ABD, the base is AD and the height is BD. For triangle ABC, we can use Heron's formula because we know all three sides.
AreaAB=12×AD×BDArea_{AB} = \frac{1}{2} \times AD \times BDAreaABC=s(sAB)(sBC)(sAC)Area_{ABC} = \sqrt{s(s - AB)(s - BC)(s - AC)}where s=AB+BC+AC2s = \frac{AB + BC + AC}{2} is the semi-perimeter of the triangle.

STEP 9

Plug in the values for AD, BD, AB, BC, and AC to calculate the areas of the triangles.
AreaAB=2×14.30×12Area_{AB} = \frac{}{2} \times14.30 \times12AreaABC=s(sAB)(sBC)(sAC)Area_{ABC} = \sqrt{s(s - AB)(s - BC)(s - AC)}

STEP 10

Calculate the areas of the triangles.
AreaAB=2×14.30×12=85.8cm2Area_{AB} = \frac{}{2} \times14.30 \times12 =85.8 \, cm^2AreaABC=s(sAB)(sBC)(sAC)=cm2Area_{ABC} = \sqrt{s(s - AB)(s - BC)(s - AC)} = \ldots \, cm^2

STEP 11

Add the areas of the triangles to find the area of the quadrilateral.
Area_{ABCD} = Area_{AB} + Area_{ABC} =85.8 \, cm^ + \ldots \, cm^ = \ldots \, cm^

STEP 12

Finally, we need to calculate the shortest distance from B to AC. This is the perpendicular distance from a point to a line in a triangle, which can be found using the formula 2×AreaABCAC\frac{2 \times Area_{ABC}}{AC}.
BD=2×AreaABCACBD_{\perp} = \frac{2 \times Area_{ABC}}{AC}

STEP 13

Plug in the values for the area of triangle ABC and the length of AC to calculate the shortest distance from B to AC.
BD=2×cm2cmBD_{\perp} = \frac{2 \times \ldots \, cm^2}{\ldots \, cm}

STEP 14

Calculate the shortest distance from B to AC.
BD=2×cm2cm=cmBD_{\perp} = \frac{2 \times \ldots \, cm^2}{\ldots \, cm} = \ldots \, cmThe area of the quadrilateral ABC is cm2\ldots \, cm^2 and the shortest distance from B to AC is cm\ldots \, cm.

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