Question

A surveyor, Toby, measures the distance between two landmarks and the point where he stands. He also measured the angles between the landmarks in degrees.
the triangle has
two sides(65,55)
angles (40,30)

What is the distance, x, between the two landmarks? Round the answer to the nearest tenth.

32.5 m
42.1 m
85.1 m
98.5 m

Answer 1

Answer:

98.5 m

Step-by-step explanation:

Refer the attached figure

AB = 55

AD = 65

∠ABC=40°

∠ADC = 30°

We are supposed to find the distance between the two landmarks i.e. BD = BC+CD

In ΔABC

$Cos \theta = \frac{Base}{Hypotenuse}$

$Cos 40^{\circ} = \frac{BC}{AB}$

$0.76604444= \frac{BC}{55}$

$0.76604444 \times 55 =BC$

$42.132442 =BC$

In ΔADC

$Cos \theta = \frac{Base}{Hypotenuse}$

$Cos 30^{\circ} = \frac{CD}{AD}$

$0.8660254= \frac{CD}{65}$

$0.8660254 \times 65 =CD$

$56.291651 =CD$

So, BD = BC+CD=42.132442+56.291651=98.424≈ 98.5

Hence the distance between the two landmarks is 98.5 m.

Answer 2

The Set Up:

x² = (Side1)² + (Side2)² - 2[(Side1)(Side2)]

Solution:

cos(Toby's Angle) • x² = 55² + 65² - 2[(55)(65)] cos(110°)

x² = 3025 + 4225 -7150[cos(110°)]

x² = 7250 - 2445.44x =

√4804.56x = 69.31m

The distance, x, between two landmarks is 69.31m.

Note: The answer choices given are incorrect.