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Natalka [10]
3 years ago
8

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, th

e angle of elevation to the top is °.
Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
4 0

Answer:

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

Step-by-step explanation:

Incomplete question:

<em></em>\angle CBD = 53^\circ<em></em>

<em></em>\angle CAB = 35^\circ<em></em>

<em></em>AB = 75<em></em>

<em></em>

<em>See attachment for complete question</em>

Required

Determine the equation to find x

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

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

\angle ABC + 53= 180

Collect like terms

\angle ABC =- 53+ 180

\angle ABC =127^\circ

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

\angle ACB + 127 + 35 = 180

Collect like terms

\angle ACB =- 127 - 35 + 180

\angle ACB =18

Apply sine rule

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

In this case:

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

This gives:

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

irakobra [83]3 years ago
4 0

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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