## Solve Lesson 21 Page 81 Math 10 – Kite >

Topic

A person standing in position AOn the roof of a house 4 m high, a observer is observing a tree 20 m tall and measuring $$\widehat {BAC} = {45^0}$$ (Figure 27). Calculate the tree’s height (round to tenths of a meter)

Solution method – See details

Step 1: Use the Pythagorean theorem to calculate the length AB byABH square at H

Step 2: Use trigonometric ratios in right triangles to calculate angle ABH then calculate the angle ABC

Step 3: Calculate the angle ACB and use the sine theorem to calculate the length BC byABC then conclude

Detailed explanation

Apply Pythagorean theorem toABH square at H we have: $$AB = \sqrt {A{H^2} + H{B^2}} = \sqrt {{4^2} + {{20}^2}} \approx 20.4$$ ( m)

ConsiderABH square at H yes $$\tan \widehat {ABH} = \frac{{AH}}{{BH}} = \frac{1}{5} \Rightarrow \widehat {ABH} \approx 11,{3^0}$$

We have: $$\widehat {ABH} + \widehat {ABC} = {90^0} \Rightarrow \widehat {ABC} = {90^0} – \widehat {ABH} = 78,{7^0}\ ) \( \Rightarrow \widehat {ACB} = {180^0} – (\widehat {ABC} + \widehat {CAB}) = 56,{3^0}$$

Apply the sine theorem toABC we have: $$\frac{{BC}}{{\sin \widehat {BAC}}} = \frac{{AB}}{{\sin \widehat {ACB}}} \Rightarrow BC = \frac{{ AB.\sin \widehat {BAC}}}{{\sin \widehat {ACB}}} = \frac{{20.4.\sin {{45}^0}}}{{\sin 56,{3 ^0}}} \approx 17.3$$ (m)

So the height of the tree is 17.3 m