# CBSE Class 10 Maths 2018 Question Paper Solution | Q29

###### CBSE Solved Question Paper 4 Mark Question 29 | Heights & Distances

Applications of Trigonometry question that appeared as a four mark question in CBSE class 10 math 2018 question paper. Heights and distances is the concept tested. Basic knowledge of using the appropriate trigonometric ratios and corresponding values is required to solve this question. A relatively easy question.

Question 29: As observed from the top of a 100 m high light house from the sea – level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3 = 1.732]

## NCERT Solution to Class 10 Maths

### Explanatory Answer | CBSE Class 10 Math 2018 Paper Question 29

Let AB be the light house.
Let C and D be the 2 ships.

∠EAD and ∠EAC are the angles depression of the two ships from the top of the light house.
AE and BD are parallel
∴ ∠EAD = ∠BDA = 30°
And ∠EAC = ∠BCA = 45°
Essentially, angle of depression is the same as the angle of elevation.

#### Steps to solve this Applications of Trigonometry Question

Step 1: Compute BC
Step 2: Compute BD
Step 3: Find CD = BD – BC

In right ΔABC tan 45° = $$frac{AB}{BC}$ 1 = $\frac{100}{BC}$ Or BC = 100 m #### Step 2: Compute length of BD In right triangle ABD tan 30° = $\frac{AB}{BD}$ $\frac{1}{√3}$ = $\frac{100}{BD}$ Or BD = 100√3 #### Step 3: Compute CD, the distance between the two ships CD = BD – BC = 100√3 – 100 = 100$√3 – 1) = 100(1.732 – 1)
= 100 × 0.732 = 73.2 m

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