Answered

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At the base of a pyramid, a surveyor determines that the angle of elevation to the top is °. At a point meters from the base, the angle of elevation to the top is °.

Sagot :

MrRoyal

Answer:

[tex]\frac{(\sin 18^\circ)}{75} = \frac{(\sin 35^\circ)}{x}[/tex]

Step-by-step explanation:

Incomplete question:

[tex]\angle CBD = 53^\circ[/tex]

[tex]\angle CAB = 35^\circ[/tex]

[tex]AB = 75[/tex]

See attachment for complete question

Required

Determine the equation to find x

First, is to complete the angles of the triangle (ABC and ACB)

[tex]\angle ABC + \angle CBD = 180[/tex] --- angle on a straight line

[tex]\angle ABC + 53= 180[/tex]

Collect like terms

[tex]\angle ABC =- 53+ 180[/tex]

[tex]\angle ABC =127^\circ[/tex]

[tex]\angle ABC + \angle ACB + \angle CAB = 180[/tex] --- angles in a triangle

[tex]\angle ACB + 127 + 35 = 180[/tex]

Collect like terms

[tex]\angle ACB =- 127 - 35 + 180[/tex]

[tex]\angle ACB =18[/tex]

Apply sine rule

[tex]\frac{\sin A}{a} = \frac{\sin B}{b}[/tex]

In this case:

[tex]\frac{\sin ACB}{AB} = \frac{\sin CAB}{x}[/tex]

This gives:

[tex]\frac{(\sin 18^\circ)}{75} = \frac{(\sin 35^\circ)}{x}[/tex]

View image MrRoyal
RheaChakraborty

Step-by-step explanation:

sin18

∘

)

=

x

(sin35

∘

)

Step-by-step explanation:

Incomplete question:

\angle CBD = 53^\circ∠CBD=53

∘

\angle CAB = 35^\circ∠CAB=35

∘

AB = 75AB=75

See attachment for complete question

Required

Determine the equation to find x

First, is to complete the angles of the triangle (ABC and ACB)

\angle ABC + \angle CBD = 180∠ABC+∠CBD=180 --- angle on a straight line

\angle ABC + 53= 180∠ABC+53=180

Collect like terms

\angle ABC =- 53+ 180∠ABC=−53+180

\angle ABC =127^\circ∠ABC=127

∘

\angle ABC + \angle ACB + \angle CAB = 180∠ABC+∠ACB+∠CAB=180 --- angles in a triangle

\angle ACB + 127 + 35 = 180∠ACB+127+35=180

Collect like terms

\angle ACB =- 127 - 35 + 180∠ACB=−127−35+180

\angle ACB =18∠ACB=18

Apply sine rule

\frac{\sin A}{a} = \frac{\sin B}{b}

a

sinA

=

b

sinB

In this case:

\frac{\sin ACB}{AB} = \frac{\sin CAB}{x}

AB

sinACB

=

x

sinCAB

This gives:

\frac{(\sin 18^\circ)}{75} = \frac{(\sin 35^\circ)}{x}

75

(sin18

∘

)

=

x

(sin35

∘

)

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