Question

In triangle ABC right angled at C, AC = 20 cm and AB – BC = 8 cm. Determine the values of sec A+tan A. *

Answer 1

Answer:

Step-by-step explanation:

Firstly it is right triangle with AC as 20 and since right angled at C AB should be hypotenuse and BC is the base.

Let BC be x so AB be x+8

Using Pythagoras theorem

Acc to ques AB^2 =AC^2+BC^2

(X+8)^2=x^2+20^2

(X+8)^2-x^2 =400

(X+8+x)(x+8-x)=400

(2x+8)8=400

2x+8= 50

X=42/2=21=BC

AB =21+8=29

Now in tge triangle ABC seca+tan A

= hypot/base+perpendicular /base ( with respect to A)

29/20+21/20

=50/20= 2.5