gracegaebel
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If someone could please help me with q's 4 and 6 that would be great. The answers are 35.5m for #4 and 890m for #6 but I'm not sure on how to solve it using sine law.

If someone could please help me with qs 4 and 6 that would be great The answers are 355m for 4 and 890m for 6 but Im not sure on how to solve it using sine law class=


Answer :

Answer:

890m

Step-by-step explanation:

I don't believe that using sine law will help, but setting up an equation does!

First let's draw a right triangle, with the vertical leg being the height of the mountain we're trying to find. They say that Cece measure the angle of elevation, or the angle opposite of the "h" side length, as 34 degrees.  The horizontal leg or the distance that Cece is away from the base of the mountain is x (they don't explicitly tell us how far).

Then, they say that David who is 365m closer measures the angle elevation as 43. So we'd need to draw another hypotenuse in the triangle we have where the angle opposite of the "h" side length is 43 degrees. Since he's closer the distance between him and Cece is 365 meaning that his distance relative to the mountain is x - 665.

Now we use our basic trigonometric functions to set up equations!

[tex]tan(34)=\frac{h}{x}[/tex]           (1)

[tex]tan(43)=\frac{h}{x-365}[/tex]          (2)

We can multiply each of the denominators on both sides of their respective equations to get the h's by itself.

xtan(34) = h     (3)                (this is (1) but just rearranged)

tan(43)*(x-365) = h     (4)      (this is (2) just rearranged)

Now we can equate (3) and (4)

xtan(34) = tan(43)*(x-365)

xtan(34) = xtan(43) - tan(43)*365     (distribute the right side)

xtan(34) - xtan(43) = -tan(43)*365    

(bring the terms with x's on one side via subtraction)

x(tan(34) - tan(43)) = -tan(43)*365       (factor)

x = [tex]\frac{-tan(43)*365}{(tan(34)-tan(43))}[/tex]      (divide the term multiplied to the x to isolate it)

x ≈ 1319.22

We're not done yet! This is only the distance between Cece and the base of the mountain! We can now plug this into either (1), (2), (3) or (4) to get h, but to make things simpler we focus on (3).

1319.22*tan(34) = 889.826 or 890m

Let me know if you have any questions!

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dashataran

Answer:

4. Solution:

Suppose:

CD = height of building

A = Position of Andrew

B = Position of Betty

Now,

In triangle BCD,

[tex]\tan 60^\circ=\dfrac{\text{CD}}{\text{CB}}\\\\\text{or, }\sqrt{3}=\dfrac{\text{CD}}{\text{CB}}\\\\\text{or, CB}=\dfrac{\text{CD}}{\sqrt3}........(1)[/tex]

In triangle ACD,

[tex]\tan 45^\circ=\dfrac{\text{CD}}{\text{AC}}\\\\\text{or, }1=\dfrac{\text{CD}}{\text{AB+CB}}\\\\\text{or, }15+\text{CB}=\text{CD}\\\\\text{or, CB}=\text{CD}-15.......(2)[/tex]

Equating equations(1) and (2),

[tex]\dfrac{\text{CD}}{\sqrt3}=\text{CD}-15\\\\\text{or, CD}=\sqrt3\text{CD}-15\sqrt3\\\\\text{or, }\sqrt3\text{CD}-\text{CD}=15\sqrt3\\\\\text{or, CD}(\sqrt{3}-1)=15\sqrt3\\\\\text{or, CD}=\dfrac{15\sqrt3}{\sqrt3-1}\\\\\text{or, CD}\approx35.5\text{m}[/tex]

So the height of the building is about 35.5m.

6. Solution:

Let:

AB = height of mountain

C = Position of Cece

D = Position of David

Now,

In triangle ABD,

[tex]\tan \text{43}^\circ=\dfrac{\text{AB}}{\text{AD}}\\\\\text{or, }\tan 43^\circ=\dfrac{\text{AB}}{\text{AD}}\\\\\text{or, AD}=\dfrac{\text{AB}}{\tan 43^\circ}\\\\\text{or, AD}=\text{AB}\cot 43^\circ......(1)[/tex]

In triangle ABC,

[tex]\tan 34^\circ=\dfrac{\text{AC}}{\text{AB}}\\\\\text{or, }\tan 34^\circ=\dfrac{\text{AB}}{\text{CD + AD}}\\\\\text{or, }\tan 34^\circ=\dfrac{\text{AB}}{365+\text{AD}}[/tex]

From equation(1),

[tex]\text{or, }\tan34^\circ=\dfrac{\text{AB}}{365+\text{AB}\cot 43^\circ}[/tex]

[tex]\text{or, }365(\tan34^\circ)+\text{AB}(\cot 43^\circ)(\tan 34^\circ)=\text{AB}\\\\\text{or, }365(\tan34^\circ)=\text{AB}-\text{AB}(\cot 43^\circ)(\tan 34^\circ)=\text{AB}\{1-(\cot 43^\circ)(\tan 34^\circ)\}\\\\\text{or, AB}=\dfrac{365(\tan 34^\circ)}{1-(\cot43^\circ)(\tan34^\circ)}\\\\\text{or, AB}\approx889.82\text{m}[/tex]

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