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Kitty [74]
3 years ago
8

Surveyors wished to know the distance across a lake between A and B. They measured the distance from B to C as 100 m and the ang

le from C to A to be 55°. Knowing that the angle between CB and BA was 90°, they used trig to solve for the distance. Show how it was done

Mathematics
1 answer:
kherson [118]3 years ago
5 0

Answer:

<h3>AB ≈ 70m</h3>

Step-by-step explanation:

Check the attachment for the diagram.

You can see from the diagram that it is a right angled triangle with opposite side AB and adjacent side BC. Using SOH, CAH, TOA trig identity to get the length of AB. According to TOA;

tan 35° = opposite/adjacent

tan 35° = AB/BC

tan 35° = AB/100

AB = 100tan35°

AB = 100 * 0.7002

AB = 70.02m

Hence the distance across a lake between A and B is approximately 70m

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Step-by-step explanation:

Given:

f(x) = - 3x - 2 and

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We require,

h(x) = g(x) - f(x)

Substituting f(x) and g(x) we get

h(x) = (-2x-4)-(-3x-2)\\\\h(x) =(-2x-4) +3x+2......Minus\ sign\ will\ go\ into\ bracket\\\\\therefore h(x)=x-2

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A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base
Olenka [21]

Answer:

The base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = x\times\frac{1}{2} =\frac{x}{2} \ m.

The height of triangle = x\times \frac{3}{4} =\frac{3x}{4}\ m.

The area of triangle = 16\ m^2.

Now, we put the formula of area to solve:

Area=\frac{1}{2} \times base\times height

16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}

16=\frac{3x^2}{16}

<em>Multiplying both sides by 16 we get:</em>

<em />256=3x^2<em />

<em>Dividing both sides by 3 we get:</em>

<em />\frac{256}{3} =x^2<em />

<em>Using square root on both sides we get:</em>

\frac{16}{\sqrt{3}}=x

x=\frac{16}{\sqrt{3}}

Now, by substituting the value of x to get the base and height:

Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.

<em>So, the base of triangle = </em>\frac{8}{\sqrt{3}} \ m.<em />

Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.

<em>Thus, the height of triangle =  </em>\frac{12}{\sqrt{3}} \ m.<em />

Therefore, the base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

7 0
4 years ago
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